any help would be appreciated. Moreover, to find the function, we need to use the given information correctly. As the distance between x and x+h gets smaller, the secant line that weve shown will approach the tangent line representing the functions derivative. Materials experience thermal strainchanges in volume or shapeas temperature changes. $(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ - Quotient Rule, $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$ - Chain Rule, $\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}\arccos(x)=-\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}\text{arccot}(x)=-\frac{1}{1+x^2}$, $\frac{d}{dx}\text{arcsec}(x)=\frac{1}{x\sqrt{x^2-1}}$, $\frac{d}{dx}\text{arccsc}(x)=-\frac{1}{x\sqrt{x^2-1}}$, Definition of a derivative # " " = lim_{h to 0} ((e^(0+h)-e^0))/{h} # The practice problem generator allows you to generate as many random exercises as you want. You can also get a better visual and understanding of the function by using our graphing tool. The corresponding change in y is written as dy. Displaying the steps of calculation is a bit more involved, because the Derivative Calculator can't completely depend on Maxima for this task. The derivative is a measure of the instantaneous rate of change which is equal to: \(f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). They are a part of differential calculus. STEP 2: Find \(\Delta y\) and \(\Delta x\). If it can be shown that the difference simplifies to zero, the task is solved. _.w/bK+~x1ZTtl Note for second-order derivatives, the notation is often used. We take the gradient of a function using any two points on the function (normally x and x+h). Solved Example on One-Sided Derivative: Is the function f(x) = |x + 7| differentiable at x = 7 ? A variable function is a polynomial function that takes the shape of a curve, so is therefore a function that has an always-changing gradient. Step 3: Click on the "Calculate" button to find the derivative of the function. $\operatorname{f}(x) \operatorname{f}'(x)$. Sign up to highlight and take notes. You can also check your answers! An extremely well-written book for students taking Calculus for the first time as well as those who need a refresher. Divide both sides by \(h\) and let \(h\) approach \(0\): \[ \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = \lim_{ h \to 0} \frac{ f\left( 1+ \frac{h}{x} \right) }{h}. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. What is the definition of the first principle of the derivative? Let us analyze the given equation. Learn more in our Calculus Fundamentals course, built by experts for you. Since \( f(1) = 0 \) \((\)put \( m=n=1 \) in the given equation\(),\) the function is \( \displaystyle \boxed{ f(x) = \text{ ln } x }. In this section, we will differentiate a function from "first principles". It is also known as the delta method. This website uses cookies to ensure you get the best experience on our website. ZL$a_A-. + x^4/(4!) Find the values of the term for f(x+h) and f(x) by identifying x and h. Simplify the expression under the limit and cancel common factors whenever possible. So actually this example was chosen to show that first principle is also used to check the "differentiability" of a such a piecewise function, which is discussed in detail in another wiki. Example Consider the straight line y = 3x + 2 shown below \]. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin . We also show a sequence of points Q1, Q2, . \sin x && x> 0. Symbolab is the best derivative calculator, solving first derivatives, second derivatives, higher order derivatives, derivative at a point, partial derivatives, implicit derivatives, derivatives using definition, and more. \[\begin{array}{l l} & = \boxed{1}. MathJax takes care of displaying it in the browser. This time we are using an exponential function. While graphing, singularities (e.g. poles) are detected and treated specially. Use parentheses! We take two points and calculate the change in y divided by the change in x. A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. For example, it is used to find local/global extrema, find inflection points, solve optimization problems and describe the motion of objects. For the next step, we need to remember the trigonometric identity: \(\sin(a + b) = \sin a \cos b + \sin b \cos a\), The formula to differentiate from first principles is found in the formula booklet and is \(f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\), More about Differentiation from First Principles, Derivatives of Inverse Trigonometric Functions, General Solution of Differential Equation, Initial Value Problem Differential Equations, Integration using Inverse Trigonometric Functions, Particular Solutions to Differential Equations, Frequency, Frequency Tables and Levels of Measurement, Absolute Value Equations and Inequalities, Addition and Subtraction of Rational Expressions, Addition, Subtraction, Multiplication and Division, Finding Maxima and Minima Using Derivatives, Multiplying and Dividing Rational Expressions, Solving Simultaneous Equations Using Matrices, Solving and Graphing Quadratic Inequalities, The Quadratic Formula and the Discriminant, Trigonometric Functions of General Angles, Confidence Interval for Population Proportion, Confidence Interval for Slope of Regression Line, Confidence Interval for the Difference of Two Means, Hypothesis Test of Two Population Proportions, Inference for Distributions of Categorical Data. Test your knowledge with gamified quizzes. 224 0 obj <>/Filter/FlateDecode/ID[<474B503CD9FE8C48A9ACE05CA21A162D>]/Index[202 43]/Info 201 0 R/Length 103/Prev 127199/Root 203 0 R/Size 245/Type/XRef/W[1 2 1]>>stream A derivative is simply a measure of the rate of change. Hope this article on the First Principles of Derivatives was informative. Nie wieder prokastinieren mit unseren Lernerinnerungen. Now we need to change factors in the equation above to simplify the limit later. Let \( m =x \) and \( n = 1 + \frac{h}{x}, \) where \(x\) and \(h\) are real numbers. Prove that #lim_(x rarr2) ( 2^x-4 ) / (x-2) =ln16#? First Derivative Calculator First Derivative Calculator full pad Examples Related Symbolab blog posts High School Math Solutions - Derivative Calculator, Logarithms & Exponents In the previous post we covered trigonometric functions derivatives (click here). Choose "Find the Derivative" from the topic selector and click to see the result! Full curriculum of exercises and videos. Given a function , there are many ways to denote the derivative of with respect to . m_+ & = \lim_{h \to 0^+} \frac{ f(0 + h) - f(0) }{h} \\ Now lets see how to find out the derivatives of the trigonometric function. > Differentiation from first principles. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. + #. \end{array} + x^3/(3!) Also, had we known that the function is differentiable, there is in fact no need to evaluate both \( m_+ \) and \( m_-\) because both have to be equal and finite and hence only one should be evaluated, whichever is easier to compute the derivative. David Scherfgen 2023 all rights reserved. This limit is not guaranteed to exist, but if it does, is said to be differentiable at . Differentiation from first principles - Calculus The Applied Maths Tutor 934 subscribers Subscribe Save 10K views 9 years ago This video tries to explain where our simplified rules for. Calculus - forum. 1.4 Derivatives 19 2 Finding derivatives of simple functions 30 2.1 Derivatives of power functions 30 2.2 Constant multiple rule 34 2.3 Sum rule 39 3 Rates of change 45 3.1 Displacement and velocity 45 3.2 Total cost and marginal cost 50 4 Finding where functions are increasing, decreasing or stationary 53 4.1 Increasing/decreasing criterion 53 UGC NET Course Online by SuperTeachers: Complete Study Material, Live Classes & More. We choose a nearby point Q and join P and Q with a straight line. The derivative of a constant is equal to zero, hence the derivative of zero is zero. The Derivative Calculator lets you calculate derivatives of functions online for free! Using the trigonometric identity, we can come up with the following formula, equivalent to the one above: \[f'(x) = \lim_{h\to 0} \frac{(\sin x \cos h + \sin h \cos x) - \sin x}{h}\]. Differentiation from First Principles. Create flashcards in notes completely automatically. When x changes from 1 to 0, y changes from 1 to 2, and so the gradient = 2 (1) 0 (1) = 3 1 = 3 No matter which pair of points we choose the value of the gradient is always 3. Let's look at another example to try and really understand the concept. For the next step, we need to remember the trigonometric identity: \(cos(a +b) = \cos a \cdot \cos b - \sin a \cdot \sin b\). We take two points and calculate the change in y divided by the change in x. Suppose \( f(x) = x^4 + ax^2 + bx \) satisfies the following two conditions: \[ \lim_{x \to 2} \frac{f(x)-f(2)}{x-2} = 4,\quad \lim_{x \to 1} \frac{f(x)-f(1)}{x^2-1} = 9.\ \]. tothebook. But when x increases from 2 to 1, y decreases from 4 to 1. \[ It transforms it into a form that is better understandable by a computer, namely a tree (see figure below). getting closer and closer to P. We see that the lines from P to each of the Qs get nearer and nearer to becoming a tangent at P as the Qs get nearer to P. The lines through P and Q approach the tangent at P when Q is very close to P. So if we calculate the gradient of one of these lines, and let the point Q approach the point P along the curve, then the gradient of the line should approach the gradient of the tangent at P, and hence the gradient of the curve. Want to know more about this Super Coaching ? Differentiation from first principles of some simple curves For any curve it is clear that if we choose two points and join them, this produces a straight line. The general notion of rate of change of a quantity \( y \) with respect to \(x\) is the change in \(y\) divided by the change in \(x\), about the point \(a\). \end{align}\]. Values of the function y = 3x + 2 are shown below. If you have any questions or ideas for improvements to the Derivative Calculator, don't hesitate to write me an e-mail. It helps you practice by showing you the full working (step by step differentiation). It is also known as the delta method. hYmo6+bNIPM@3ADmy6HR5 qx=v! ))RA"$# We simply use the formula and cancel out an h from the numerator. Pick two points x and \(x+h\). We have a special symbol for the phrase. We will have a closer look to the step-by-step process below: STEP 1: Let \(y = f(x)\) be a function. + } #, # \ \ \ \ \ \ \ \ \ = 0 +1 + (2x)/(2!) \lim_{h \to 0} \frac{ f(4h) + f(2h) + f(h) + f\big(\frac{h}{2}\big) + f\big(\frac{h}{4}\big) + f\big(\frac{h}{8}\big) + \cdots }{h} = & f'(0) \times 8\\ Such functions must be checked for continuity first and then for differentiability. Sign up to read all wikis and quizzes in math, science, and engineering topics. Differentiation from First Principles The formal technique for finding the gradient of a tangent is known as Differentiation from First Principles. Follow the below steps to find the derivative of any function using the first principle: Learnderivatives of cos x,derivatives of sin x,derivatives of xsinxandderivative of 2x, A generalization of the concept of a derivative, in which the ordinary limit is replaced by a one-sided limit. Pick two points x and x + h. STEP 2: Find \(\Delta y\) and \(\Delta x\). 1. A function satisfies the following equation: \[ \lim_{h \to 0} \frac{ f(4h) + f(2h) + f(h) + f\big(\frac{h}{2}\big) + f\big(\frac{h}{4}\big) + f\big(\frac{h}{8}\big) + \cdots}{h} = 64. Now this probably makes the next steps not only obvious but also easy: \[ \begin{align} Maybe it is not so clear now, but just let us write the derivative of \(f\) at \(0\) using first principle: \[\begin{align} Enter the function you want to differentiate into the Derivative Calculator. For example, this involves writing trigonometric/hyperbolic functions in their exponential forms. For example, the lattice parameters of elemental cesium, the material with the largest coefficient of thermal expansion in the CRC Handbook, 1 change by less than 3% over a temperature range of 100 K. . Use parentheses, if necessary, e.g. "a/(b+c)". In each calculation step, one differentiation operation is carried out or rewritten. This expression is the foundation for the rest of differential calculus: every rule, identity, and fact follows from this. \frac{\text{d}}{\text{d}x} f(x) & = \lim_{h \to 0} \frac{ f(a + h) - f(a) }{h} \\ \(m_{tangent}=\lim _{h{\rightarrow}0}{y\over{x}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). Determine, from first principles, the gradient function for the curve : f x x x( )= 2 2 and calculate its value at x = 3 ( ) ( ) ( ) 0 lim , 0 h f x h f x fx h An expression involving the derivative at \( x=1 \) is most likely to come when we differentiate the given expression and put one of the variables to be equal to one. %PDF-1.5 % The Derivative Calculator has to detect these cases and insert the multiplication sign. Stop procrastinating with our smart planner features. Differentiate from first principles \(y = f(x) = x^3\). both exists and is equal to unity. We can take the gradient of PQ as an approximation to the gradient of the tangent at P, and hence the rate of change of y with respect to x at the point P. The gradient of PQ will be a better approximation if we take Q closer to P. The table below shows the effect of reducing PR successively, and recalculating the gradient. Practice math and science questions on the Brilliant iOS app. Wolfram|Alpha is a great calculator for first, second and third derivatives; derivatives at a point; and partial derivatives. Think about this limit for a moment and we can rewrite it as: #lim_{h to 0} ((e^h-1))/{h} = lim_{h to 0} ((e^h-e^0))/{h} # The "Checkanswer" feature has to solve the difficult task of determining whether two mathematical expressions are equivalent. For this, you'll need to recognise formulas that you can easily resolve. Make sure that it shows exactly what you want. They are also useful to find Definite Integral by Parts, Exponential Function, Trigonometric Functions, etc. Acceleration is the second derivative of the position function. \]. Follow the following steps to find the derivative by the first principle. Create and find flashcards in record time. Loading please wait!This will take a few seconds. Just for the sake of curiosity, I propose another way to calculate the derivative of f: f ( x) = 1 x 2 ln f ( x) = ln ( x 2) 2 f ( x) f ( x) = 1 2 ( x 2) f ( x) = 1 2 ( x 2) 3 / 2. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. The sign of the second derivative tells us whether the slope of the tangent line to f is increasing or decreasing. New Resources. Forgot password? We want to measure the rate of change of a function \( y = f(x) \) with respect to its variable \( x \). The graph of y = x2. Earn points, unlock badges and level up while studying. The rate of change at a point P is defined to be the gradient of the tangent at P. NOTE: The gradient of a curve y = f(x) at a given point is defined to be the gradient of the tangent at that point. Calculating the rate of change at a point = &64. Both \(f_{-}(a)\text{ and }f_{+}(a)\) must exist. Everything you need for your studies in one place. 0 && x = 0 \\ 202 0 obj <> endobj Click the blue arrow to submit. + x^3/(3!) This means using standard Straight Line Graphs methods of \(\frac{\Delta y}{\Delta x}\) to find the gradient of a function. button is clicked, the Derivative Calculator sends the mathematical function and the settings (differentiation variable and order) to the server, where it is analyzed again. & = \lim_{h \to 0} \frac{ \sin h}{h} \\ However, although small, the presence of . If you don't know how, you can find instructions. \end{cases}\], So, using the terminologies in the wiki, we can write, \[\begin{align} # " " = f'(0) # (by the derivative definition). \[\begin{align} The gesture control is implemented using Hammer.js. Wolfram|Alpha calls Wolfram Languages's D function, which uses a table of identities much larger than one would find in a standard calculus textbook. The Derivative Calculator lets you calculate derivatives of functions online for free! [9KP ,KL:]!l`*Xyj`wp]H9D:Z nO V%(DbTe&Q=klyA7y]mjj\-_E]QLkE(mmMn!#zFs:StN4%]]nhM-BR' ~v bnk[a]Rp`$"^&rs9Ozn>/`3s @ For each calculated derivative, the LaTeX representations of the resulting mathematical expressions are tagged in the HTML code so that highlighting is possible. Then as \( h \to 0 , t \to 0 \), and therefore the given limit becomes \( \lim_{t \to 0}\frac{nf(t)}{t} = n \lim_{t \to 0}\frac{f(t)}{t},\) which is nothing but \( n f'(0) \). How to get Derivatives using First Principles: Calculus - YouTube 0:00 / 8:23 How to get Derivatives using First Principles: Calculus Mindset 226K subscribers Subscribe 1.7K 173K views 8. In "Options" you can set the differentiation variable and the order (first, second, derivative). We say that the rate of change of y with respect to x is 3. \end{align} \], Therefore, the value of \(f'(0) \) is 8. Question: Using differentiation from first principles only, determine the derivative of y=3x^(2)+15x-4 You're welcome to make a donation via PayPal. 0 & = \lim_{h \to 0} (2+h) \\ endstream endobj startxref & = \boxed{0}. Not what you mean? As h gets small, point B gets closer to point A, and the line joining the two gets closer to the REAL tangent at point A. & = \lim_{h \to 0} \frac{ h^2}{h} \\ Q is a nearby point. I am having trouble with this problem because I am unsure what to do when I have put my function of f (x+h) into the . Learn more about: Derivatives Tips for entering queries Enter your queries using plain English. Often, the limit is also expressed as \(\frac{\text{d}}{\text{d}x} f(x) = \lim_{x \to c} \frac{ f(x) - f(c) }{x-c} \). \end{array} & = \lim_{h \to 0} \frac{ (1 + h)^2 - (1)^2 }{h} \\ STEP 1: Let y = f(x) be a function. > Differentiating sines and cosines. How can I find the derivative of #y=c^x# using first principles, where c is an integer? & = \lim_{h \to 0} \frac{ \binom{n}{1}2^{n-1}\cdot h +\binom{n}{2}2^{n-2}\cdot h^2 + \cdots + h^n }{h} \\ This special exponential function with Euler's number, #e#, is the only function that remains unchanged when differentiated. \end{align}\]. How do we differentiate a trigonometric function from first principles? It means either way we have to use first principle! \end{array}\]. At first glance, the question does not seem to involve first principle at all and is merely about properties of limits. We write. Example: The derivative of a displacement function is velocity. Now, for \( f(0+h) \) where \( h \) is a small negative number, we would use the function defined for \( x < 0 \) since \(h\) is negative and hence the equation. So for a given value of \( \delta \) the rate of change from \( c\) to \( c + \delta \) can be given as, \[ m = \frac{ f(c + \delta) - f(c) }{(c+ \delta ) - c }.\]. \(\Delta y = e^{x+h} -e^x = e^xe^h-e^x = e^x(e^h-1)\)\(\Delta x = (x+h) - x= h\), \(\frac{\Delta y}{\Delta x} = \frac{e^x(e^h-1)}{h}\). The derivatives are used to find solutions to differential equations. The gradient of the line PQ, QR/PR seems to approach 6 as Q approaches P. Observe that as Q gets closer to P the gradient of PQ seems to be getting nearer and nearer to 6. The derivative is a measure of the instantaneous rate of change which is equal to: f ( x) = d y d x = lim h 0 f ( x + h) - f ( x) h \(3x^2\) however the entire proof is a differentiation from first principles. Basic differentiation rules Learn Proof of the constant derivative rule By taking two points on the curve that lie very closely together, the straight line between them will have approximately the same gradient as the tangent there. Moving the mouse over it shows the text. The parser is implemented in JavaScript, based on the Shunting-yard algorithm, and can run directly in the browser. \]. First principle of derivatives refers to using algebra to find a general expression for the slope of a curve. A function \(f\) satisfies the following relation: \[ f(mn) = f(m)+f(n) \quad \forall m,n \in \mathbb{R}^{+} .\]. Exploring the gradient of a function using a scientific calculator just got easier. & = \lim_{h \to 0} \frac{ 1 + 2h +h^2 - 1 }{h} \\ Our calculator allows you to check your solutions to calculus exercises. & = \lim_{h \to 0} \frac{ f( h) - (0) }{h} \\ Evaluate the derivative of \(\sin x \) at \( x=a\) using first principle, where \( a \in \mathbb{R} \). Derivative by first principle is often used in cases where limits involving an unknown function are to be determined and sometimes the function itself is to be determined. Here are some examples illustrating how to ask for a derivative. Learn about Differentiation and Integration and Derivative of Sin 2x, \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=sinx\\ f(x+h)=sin(x+h)\\ f(x+h)f(x)= sin(x+h) sin(x) = sinxcosh + cosxsinh sinx\\ = sinx(cosh-1) + cosxsinh\\ {f(x+h) f(x)\over{h}}={ sinx(cosh-1) + cosxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sinx(cosh-1) + cosxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinx(cosh-1)\over{h}} + \lim _{h{\rightarrow}0} {cosxsinh\over{h}}\\ = sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} + cosx \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \text{Put h = 0 in first limit}\\ sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = sinx\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {cosh\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \times1 = cosx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx \end{matrix}\), \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=sinx\\ f(x+h)=sin(x+h)\\ f(x+h)f(x)= sin(x+h) sin(x) = {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0}2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{{h\over{2}}}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}})\times1\\ {\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}) = cosx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx \end{matrix}\), Learn about Derivative of Log x and Derivative of Sec Square x, \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=cosx\\ f(x+h)=cos(x+h)\\ f(x+h)f(x)= cos(x+h) cos(x) = cosxcosh sinxsinh cosx\\ = cosx(cosh-1) sinxsinh\\ {f(x+h) f(x)\over{h}}={ cosx(cosh-1) sinxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { cosx(cosh-1) sinxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosx(cosh-1)\over{h}} \lim _{h{\rightarrow}0} {sinxsinh\over{h}}\\ = cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} sinx \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \text{Put h = 0 in first limit}\\ cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = cosx\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {cosh\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \times1 = -sinx\\ f(x)={dy\over{dx}} = {d(cosx)\over{dx}} = -sinx \end{matrix}\), \(\begin{matrix}\ f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=cosx\\ f(x+h)=cos(x+h)\\ f(x+h)f(x)= cos(x+h) cos(x) = {-2sin({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {-2sin({2x+h\over{2}})sin({h\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2cos(x+{h\over{2}}){sin({h\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0}-2sin(x+{h\over{2}}){sin({h\over{2}})\over{{h\over{2}}}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}})\times1\\ {\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}}) = -sinx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = -sinx \end{matrix}\), If f(x) = tanx , find f(x) \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=tanx\\ f(x+h)=tan(x+h)\\ f(x+h)f(x)= tan(x+h) tan(x) = {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\\ {f(x+h) f(x)\over{h}}={ {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosxsin(x+h) sinxcos(x+h)\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {{sin(2x+h)+sinh\over{2}} {sin(2x+h)-sinh\over{2}}\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {1\over{cosxcos(x+h)}}\\ =1\times{1\over{cosx\times{cosx}}}\\ ={1\over{cos^2x}}\\ ={sec^2x}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = {sec^2x}\\ f(x)={dy\over{dx}} = {d(tanx)\over{dx}} = {sec^2x} \end{matrix}\), \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=sin5x\\ f(x+h)=sin(5x+5h)\\ f(x+h)f(x)= sin(5x+5h) sin(5x) = sin5xcos5h + cos5xsin5h sin5x\\ = sin5x(cos5h-1) + cos5xsin5h\\ {f(x+h) f(x)\over{h}}={ sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sin5x(cos5h-1)\over{h}} + \lim _{h{\rightarrow}0} {cos5xsin5h\over{h}}\\ = sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} + cos5x \lim _{h{\rightarrow}0} {sin5h\over{h}}\\ \text{Put h = 0 in first limit}\\ sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} = sin5x\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} 5\times{{d\over{dh}}sin5h\over{{d\over{dh}}5h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} {5cos5h\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \times5 = 5cos5x \end{matrix}\).
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Articles D