Hysteria; All Lights and Lights Out (pdf) Lights Out up to 20x20 It means either way we have to use first principle! This means we will start from scratch and use algebra to find a general expression for the slope of a curve, at any value x. The rate of change of y with respect to x is not a constant. For \( f(0+h) \) where \( h \) is a small positive number, we would use the function defined for \( x > 0 \) since \(h\) is positive and hence the equation. = & 4 f'(0) + 2 f'(0) + f'(0) + \frac{1}{2} f'(0) + \cdots \\ The sign of the second derivative tells us whether the slope of the tangent line to f is increasing or decreasing. 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. Create the most beautiful study materials using our templates. I know the derivative of x^3 should be 3x^2 from the power rule however when trying to differentiate using first principles (f'(x)=limh->0 [f(x+h)-f(x)]/h) I ended up with 3x^2+3x. + #, Differentiating Exponential Functions with Calculators, Differentiating Exponential Functions with Base e, Differentiating Exponential Functions with Other Bases. \]. sF1MOgSwEyw1zVt'B0zyn_'sim|U.^LV\#.=F?uS;0iO? But wait, \( m_+ \neq m_- \)!! The x coordinate of Q is x + dx where dx is the symbol we use for a small change, or small increment in x. Maxima takes care of actually computing the derivative of the mathematical function. & = \lim_{h \to 0}\left[ \sin a \bigg( \frac{\cos h-1 }{h} \bigg) + \cos a \bigg( \frac{\sin h }{h} \bigg)\right] \\ How can I find the derivative of #y=e^x# from first principles? At a point , the derivative is defined to be . Sign up, Existing user? Learn what derivatives are and how Wolfram|Alpha calculates them. The gesture control is implemented using Hammer.js. Follow the following steps to find the derivative of any function. Function Commands: * is multiplication oo is \displaystyle \infty pi is \displaystyle \pi x^2 is x 2 sqrt (x) is \displaystyle \sqrt {x} x Differentiation from First Principles. f'(0) & = \lim_{h \to 0} \frac{ f(0 + h) - f(0) }{h} \\ If you are dealing with compound functions, use the chain rule. For this, you'll need to recognise formulas that you can easily resolve. We have a special symbol for the phrase. However, although small, the presence of . Often, the limit is also expressed as \(\frac{\text{d}}{\text{d}x} f(x) = \lim_{x \to c} \frac{ f(x) - f(c) }{x-c} \). So the coordinates of Q are (x + dx, y + dy). Observe that the gradient of the straight line is the same as the rate of change of y with respect to x. Conic Sections: Parabola and Focus. & = 2.\ _\square \\ The "Checkanswer" feature has to solve the difficult task of determining whether two mathematical expressions are equivalent. It helps you practice by showing you the full working (step by step differentiation). Wolfram|Alpha is a great calculator for first, second and third derivatives; derivatives at a point; and partial derivatives. Let \( c \in (a,b) \) be the number at which the rate of change is to be measured. Knowing these values we can calculate the change in y divided by the change in x and hence the gradient of the line PQ. An extremely well-written book for students taking Calculus for the first time as well as those who need a refresher. It has reduced by 5 units. Differentiating functions is not an easy task! We can calculate the gradient of this line as follows. \]. & = \boxed{1}. . STEP 1: Let \(y = f(x)\) be a function. It is also known as the delta method. * 2) + (4x^3)/(3! Here are some examples illustrating how to ask for a derivative. This website uses cookies to ensure you get the best experience on our website. Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. By registering you get free access to our website and app (available on desktop AND mobile) which will help you to super-charge your learning process. 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. How do we differentiate a quadratic from first principles? (Total for question 2 is 5 marks) 3 Prove, from first principles, that the derivative of 2x3 is 6x2. \(\begin{matrix} f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{f(-7+h)f(-7)\over{h}}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{|(-7+h)+7|-0\over{h}}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{|h|\over{h}}\\ \text{as h < 0 in this case}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{-h\over{h}}\\ f_{-}(-7)=-1\\ \text{On the other hand}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{f(-7+h)f(-7)\over{h}}\\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{|(-7+h)+7|-0\over{h}}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{|h|\over{h}}\\ \text{as h > 0 in this case}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{h\over{h}}\\ f_{+}(-7)=1\\ \therefore{f_{-}(a)\neq{f_{+}(a)}} \end{matrix}\), Therefore, f(x) it is not differentiable at x = 7, Learn about Derivative of Cos3x and Derivative of Root x. Differentiating a linear function A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. Solved Example on One-Sided Derivative: Is the function f(x) = |x + 7| differentiable at x = 7 ? Now this probably makes the next steps not only obvious but also easy: \[ \begin{align} # " " = lim_{h to 0} {e^xe^h-e^(x)}/{h} # The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. In the case of taking a derivative with respect to a function of a real variable, differentiating f ( x) = 1 / x is fairly straightforward by using ordinary algebra. Let \( 0 < \delta < \epsilon \) . You can also get a better visual and understanding of the function by using our graphing tool. The x coordinate of Q is then 3.1 and its y coordinate is 3.12. We take two points and calculate the change in y divided by the change in x. This hints that there might be some connection with each of the terms in the given equation with \( f'(0).\) Let us consider the limit \( \lim_{h \to 0}\frac{f(nh)}{h} \), where \( n \in \mathbb{R}. The Derivative Calculator lets you calculate derivatives of functions online for free! Differentiation from First Principles. \]. P is the point (3, 9). Learn differential calculus for freelimits, continuity, derivatives, and derivative applications. (Total for question 4 is 4 marks) 5 Prove, from first principles, that the derivative of kx3 is 3kx2. & = \lim_{h \to 0} \frac{ \sin (a + h) - \sin (a) }{h} \\ Rate of change \((m)\) is given by \( \frac{f(x_2) - f(x_1)}{x_2 - x_1} \). Free linear first order differential equations calculator - solve ordinary linear first order differential equations step-by-step. How do we differentiate a trigonometric function from first principles? The Derivative Calculator supports solving first, second., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. We know that, \(f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). How Does Derivative Calculator Work? \[ The graph of y = x2. If the one-sided derivatives are equal, then the function has an ordinary derivative at x_o. hYmo6+bNIPM@3ADmy6HR5 qx=v! ))RA"$# Let us analyze the given equation. Did this calculator prove helpful to you? would the 3xh^2 term not become 3x when the limit is taken out? How do we differentiate from first principles? + x^3/(3!) The final expression is just \(\frac{1}{x} \) times the derivative at 1 \(\big(\)by using the substitution \( t = \frac{h}{x}\big) \), which is given to be existing, implying that \( f'(x) \) exists. Now we need to change factors in the equation above to simplify the limit later. Joining different pairs of points on a curve produces lines with different gradients. Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. Tutorials in differentiating logs and exponentials, sines and cosines, and 3 key rules explained, providing excellent reference material for undergraduate study. + (5x^4)/(5!) $\operatorname{f}(x) \operatorname{f}'(x)$. 0 && x = 0 \\ If you have any questions or ideas for improvements to the Derivative Calculator, don't hesitate to write me an e-mail. As follows: f ( x) = lim h 0 1 x + h 1 x h = lim h 0 x ( x + h) ( x + h) x h = lim h 0 1 x ( x + h) = 1 x 2. It is also known as the delta method. Note for second-order derivatives, the notation is often used. We use this definition to calculate the gradient at any particular point. The derivatives are used to find solutions to differential equations. It can be the rate of change of distance with respect to time or the temperature with respect to distance. New user? tothebook. 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. Calculating the gradient between points A & B is not too hard, and if we let h -> 0 we will be calculating the true gradient. Either we must prove it or establish a relation similar to \( f'(1) \) from the given relation. & = \lim_{h \to 0} \frac{ (1 + h)^2 - (1)^2 }{h} \\ So, the answer is that \( f'(0) \) does not exist. It implies the derivative of the function at \(0\) does not exist at all!! \(m_{tangent}=\lim _{h{\rightarrow}0}{y\over{x}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). This is also known as the first derivative of the function. The derivative of a constant is equal to zero, hence the derivative of zero is zero. Derivation of sin x: = cos xDerivative of cos x: = -sin xDerivative of tan x: = sec^2xDerivative of cot x: = -cosec^2xDerivative of sec x: = sec x.tan xDerivative of cosec x: = -cosec x.cot x. Stop procrastinating with our smart planner features. Example Consider the straight line y = 3x + 2 shown below Will you pass the quiz? The derivative is a measure of the instantaneous rate of change which is equal to: \(f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). 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. Example : We shall perform the calculation for the curve y = x2 at the point, P, where x = 3. 1. & = n2^{n-1}.\ _\square Clicking an example enters it into the Derivative Calculator. 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. Evaluate the resulting expressions limit as h0. . \[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]. We will have a closer look to the step-by-step process below: STEP 1: Let \(y = f(x)\) be a function. Consider a function \(f : [a,b] \rightarrow \mathbb{R}, \) where \( a, b \in \mathbb{R} \). 0 Let's try it out with an easy example; f (x) = x 2. We often use function notation y = f(x). Q is a nearby point. 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. First principles is also known as "delta method", since many texts use x (for "change in x) and y (for . Enter the function you want to differentiate into the Derivative Calculator. So, the change in y, that is dy is f(x + dx) f(x). [9KP ,KL:]!l`*Xyj`wp]H9D:Z nO V%(DbTe&Q=klyA7y]mjj\-_E]QLkE(mmMn!#zFs:StN4%]]nhM-BR' ~v bnk[a]Rp`$"^&rs9Ozn>/`3s @ They are a part of differential calculus. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. Calculating the rate of change at a point ZL$a_A-. The tangent line is the result of secant lines having a distance between x and x+h that are significantly small and where h0. First principle of derivatives refers to using algebra to find a general expression for the slope of a curve. \frac{\text{d}}{\text{d}x} f(x) & = \lim_{h \to 0} \frac{ f(a + h) - f(a) }{h} \\ 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. \frac{\text{d}}{\text{d}x} f(x) & = \lim_{h \to 0} \frac{ f(1 + h) - f(1) }{h} \\ %PDF-1.5 % + x^4/(4!) For example, this involves writing trigonometric/hyperbolic functions in their exponential forms. # " " = lim_{h to 0} e^x((e^h-1))/{h} # & = \lim_{h \to 0^+} \frac{ \sin (0 + h) - (0) }{h} \\ It helps you practice by showing you the full working (step by step differentiation). You find some configuration options and a proposed problem below. 202 0 obj <> endobj What is the differentiation from the first principles formula? Using the trigonometric identity, we can come up with the following formula, equivalent to the one above: \[f'(x) = \lim_{h\to 0} \frac{(\sin x \cos h + \sin h \cos x) - \sin x}{h}\]. (See Functional Equations. These changes are usually quite small, as Fig. Simplifying and taking the limit, the derivative is found to be \frac{1}{2\sqrt{x}}. Ltd.: All rights reserved. Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persnlichen Lernstatistiken. Differentiation from First Principles The First Principles technique is something of a brute-force method for calculating a derivative - the technique explains how the idea of differentiation first came to being. Look at the table of values and note that for every unit increase in x we always get an increase of 3 units in y. & = \boxed{0}. \begin{array}{l l} multipliers and divisors), derive each component separately, carefully set the rule formula, and simplify. As an Amazon Associate I earn from qualifying purchases. Please enable JavaScript. + x^3/(3!) Figure 2. This means using standard Straight Line Graphs methods of \(\frac{\Delta y}{\Delta x}\) to find the gradient of a function. Derivative by the first principle is also known as the delta method. StudySmarter is commited to creating, free, high quality explainations, opening education to all. 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. Sign up to read all wikis and quizzes in math, science, and engineering topics. Thank you! Well, in reality, it does involve a simple property of limits but the crux is the application of first principle. This is the first chapter from the whole textbook, where I would like to bring you up to speed with the most important calculus techniques as taught and widely used in colleges and at . For the next step, we need to remember the trigonometric identity: \(cos(a +b) = \cos a \cdot \cos b - \sin a \cdot \sin b\). U)dFQPQK$T8D*IRu"G?/t4|%}_|IOG$NF\.aS76o:j{ Like any computer algebra system, it applies a number of rules to simplify the function and calculate the derivatives according to the commonly known differentiation rules. Answer: d dx ex = ex Explanation: We seek: d dx ex Method 1 - Using the limit definition: f '(x) = lim h0 f (x + h) f (x) h We have: f '(x) = lim h0 ex+h ex h = lim h0 exeh ex h This special exponential function with Euler's number, #e#, is the only function that remains unchanged when differentiated. Velocity is the first derivative of the position function. For different pairs of points we will get different lines, with very different gradients. 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}. Consider the graph below which shows a fixed point P on a curve. Differentiation from First Principles Calculus Absolute Maxima and Minima Absolute and Conditional Convergence Accumulation Function Accumulation Problems Algebraic Functions Alternating Series Antiderivatives Application of Derivatives Approximating Areas Arc Length of a Curve Area Between Two Curves Arithmetic Series Average Value of a Function 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. The Derivative Calculator supports solving first, second.., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. \[\begin{align} Maxima's output is transformed to LaTeX again and is then presented to the user. Create flashcards in notes completely automatically. Derivative of a function is a concept in mathematicsof real variable that measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). STEP 2: Find \(\Delta y\) and \(\Delta x\). This time, the function gets transformed into a form that can be understood by the computer algebra system Maxima. Then we can differentiate term by term using the power rule: # d/dx e^x = d/dx{1 +x + x^2/(2!) 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. & = \sin a \lim_{h \to 0} \bigg( \frac{\cos h-1 }{h} \bigg) + \cos a \lim_{h \to 0} \bigg( \frac{\sin h }{h} \bigg) \\ Calculus Differentiating Exponential Functions From First Principles Key Questions How can I find the derivative of y = ex from first principles? m_+ & = \lim_{h \to 0^+} \frac{ f(0 + h) - f(0) }{h} \\ endstream endobj startxref any help would be appreciated. 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 . DHNR@ R$= hMhNM Please ensure that your password is at least 8 characters and contains each of the following: You'll be able to enter math problems once our session is over. What is the second principle of the derivative? The point A is at x=3 (originally, but it can be moved!) This book makes you realize that Calculus isn't that tough after all. When the "Go!" _.w/bK+~x1ZTtl Pick two points x and x + h. Coordinates are \((x, x^3)\) and \((x+h, (x+h)^3)\). At first glance, the question does not seem to involve first principle at all and is merely about properties of limits. As an example, if , then and then we can compute : . For more about how to use the Derivative Calculator, go to "Help" or take a look at the examples. In "Options" you can set the differentiation variable and the order (first, second, derivative). The rules of differentiation (product rule, quotient rule, chain rule, ) have been implemented in JavaScript code. Differentiation is the process of finding the gradient of a variable function. 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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