The Derivative Calculator will show you a graphical version of your input while you type. hYmo6+bNIPM@3ADmy6HR5 qx=v! ))RA"$# We choose a nearby point Q and join P and Q with a straight line. 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. 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. . In fact, all the standard derivatives and rules are derived using first principle. Wolfram|Alpha doesn't run without JavaScript. 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. Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. \) \(_\square\), Note: If we were not given that the function is differentiable at 0, then we cannot conclude that \(f(x) = cx \). It means either way we have to use first principle! Create flashcards in notes completely automatically. What is the differentiation from the first principles formula? It uses well-known rules such as the linearity of the derivative, product rule, power rule, chain rule and so on. The derivative is an important tool in calculus that represents an infinitesimal change in a function with respect to one of its variables. Sign up to highlight and take notes. Suppose we want to differentiate the function f(x) = 1/x from first principles. Derivative by the first principle is also known as the delta method. & = \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) \\ # " " = f'(0) # (by the derivative definition). We will have a closer look to the step-by-step process below: STEP 1: Let \(y = f(x)\) be a function. 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}. 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. There is also a table of derivative functions for the trigonometric functions and the square root, logarithm and exponential function. \[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\]. Analyzing functions Calculator-active practice: Analyzing functions . Sign up to read all wikis and quizzes in math, science, and engineering topics. Evaluate the resulting expressions limit as h0. Learn differential calculus for freelimits, continuity, derivatives, and derivative applications. Now lets see how to find out the derivatives of the trigonometric function. STEP 1: Let \(y = f(x)\) be a function. 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. So, the answer is that \( f'(0) \) does not exist. Function Commands: * is multiplication oo is \displaystyle \infty pi is \displaystyle \pi x^2 is x 2 sqrt (x) is \displaystyle \sqrt {x} x But wait, we actually do not know the differentiability of the 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. When you're done entering your function, click "Go! This is the fundamental definition of derivatives. Paid link. For any curve it is clear that if we choose two points and join them, this produces a straight line. \begin{cases} \frac{\text{d}}{\text{d}x} f(x) & = \lim_{h \to 0} \frac{ f(1 + h) - f(1) }{h} \\ Differentiation from First Principles The formal technique for finding the gradient of a tangent is known as Differentiation from First Principles. Additionly, the number #2.718281 #, which we call Euler's number) denoted by #e# is extremely important in mathematics, and is in fact an irrational number (like #pi# and #sqrt(2)#. It can be the rate of change of distance with respect to time or the temperature with respect to distance. Plugging \sqrt{x} into the definition of the derivative, we multiply the numerator and denominator by the conjugate of the numerator, \sqrt{x+h}+\sqrt{x}. Let \( t=nh \). \end{array} \[\begin{align} It is also known as the delta method. A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. Additionally, D uses lesser-known rules to calculate the derivative of a wide array of special functions. \(f(a)=f_{-}(a)=f_{+}(a)\). Pick two points x and x + h. Coordinates are \((x, x^3)\) and \((x+h, (x+h)^3)\). Learn more about: Derivatives Tips for entering queries Enter your queries using plain English. Differentiate from first principles \(y = f(x) = x^3\). In other words, y increases as a rate of 3 units, for every unit increase in x. Such functions must be checked for continuity first and then for differentiability. 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 = &64. Enter the function you want to differentiate into the Derivative Calculator. tothebook. Is velocity the first or second derivative? Now we need to change factors in the equation above to simplify the limit later. Have all your study materials in one place. Let \( m =x \) and \( n = 1 + \frac{h}{x}, \) where \(x\) and \(h\) are real numbers. 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. Hope this article on the First Principles of Derivatives was informative. You can also check your answers! We say that the rate of change of y with respect to x is 3. Their difference is computed and simplified as far as possible using Maxima. This time we are using an exponential function. Differentiation is the process of finding the gradient of a variable function. Because we are considering the graph of y = x2, we know that y + dy = (x + dx)2. + x^3/(3!) It helps you practice by showing you the full working (step by step differentiation). Observe that the gradient of the straight line is the same as the rate of change of y with respect to x. The Derivative Calculator supports solving first, second.., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. Hence, \( f'(x) = \frac{p}{x} \). & = \lim_{h \to 0} \frac{ f(h)}{h}. What is the definition of the first principle of the derivative? 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. StudySmarter is commited to creating, free, high quality explainations, opening education to all. Make sure that it shows exactly what you want. + x^4/(4!) So, the change in y, that is dy is f(x + dx) f(x). How do we differentiate a trigonometric function from first principles? Interactive graphs/plots help visualize and better understand the functions. Prove that #lim_(x rarr2) ( 2^x-4 ) / (x-2) =ln16#? This should leave us with a linear function. The x coordinate of Q is then 3.1 and its y coordinate is 3.12. %PDF-1.5 % First Principle of Derivatives refers to using algebra to find a general expression for the slope of a curve. The derivative is a measure of the instantaneous rate of change, which is equal to f' (x) = \lim_ {h \rightarrow 0 } \frac { f (x+h) - f (x) } { h } . # e^x = 1 +x + x^2/(2!) & = \lim_{h \to 0} \frac{ (1 + h)^2 - (1)^2 }{h} \\ * 2) + (4x^3)/(3! In this section, we will differentiate a function from "first principles". The limit \( \lim_{h \to 0} \frac{ f(c + h) - f(c) }{h} \), if it exists (by conforming to the conditions above), is the derivative of \(f\) at \(c\) and the method of finding the derivative by such a limit is called derivative by first principle. So differentiation can be seen as taking a limit of a gradient between two points of a function. (See Functional Equations. If you know some standard derivatives like those of \(x^n\) and \(\sin x,\) you could just realize that the above-obtained values are just the values of the derivatives at \(x=2\) and \(x=a,\) respectively. Derivative Calculator First Derivative Calculator (Solver) with Steps Free derivatives calculator (solver) that gets the detailed solution of the first derivative of a function. $(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ - 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 The derivatives are used to find solutions to differential equations. & = \lim_{h \to 0} \frac{ f( h) - (0) }{h} \\ m_+ & = \lim_{h \to 0^+} \frac{ f(0 + h) - f(0) }{h} \\ \]. \]. The most common ways are and . 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. 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. The sign of the second derivative tells us whether the slope of the tangent line to f is increasing or decreasing. Forgot password? U)dFQPQK$T8D*IRu"G?/t4|%}_|IOG$NF\.aS76o:j{ & = \lim_{h \to 0} \frac{ \sin a \cos h + \cos a \sin h - \sin a }{h} \\ Then, This is the definition, for any function y = f(x), of the derivative, dy/dx, NOTE: Given y = f(x), its derivative, or rate of change of y with respect to x is defined as. lim stands for limit and we say that the limit, as x tends to zero, of 2x+dx is 2x. The point A is at x=3 (originally, but it can be moved!) Differentiation from first principles. \frac{\text{d}}{\text{d}x} f(x) & = \lim_{h \to 0} \frac{ f(a + h) - f(a) }{h} \\ + } #, # \ \ \ \ \ \ \ \ \ = 0 +1 + (2x)/(2!) Upload unlimited documents and save them online. Clicking an example enters it into the Derivative Calculator. UGC NET Course Online by SuperTeachers: Complete Study Material, Live Classes & More. 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. So the coordinates of Q are (x + dx, y + dy). ZL$a_A-. A bit of history of calculus, with a formula you need to learn off for the test.Subscribe to our YouTube channel: http://goo.gl/s9AmD6This video is brought t. Values of the function y = 3x + 2 are shown below. m_- & = \lim_{h \to 0^-} \frac{ f(0 + h) - f(0) }{h} \\ + (4x^3)/(4!) \end{cases}\], So, using the terminologies in the wiki, we can write, \[\begin{align} The derivative can also be represented as f(x) as either f(x) or y. Using the trigonometric identity, we can come up with the following formula, equivalent to the one above: \[f'(x) = \lim_{h\to 0} \frac{(\sin x \cos h + \sin h \cos x) - \sin x}{h}\]. _.w/bK+~x1ZTtl The Derivative Calculator lets you calculate derivatives of functions online for free! At a point , the derivative is defined to be . Calculus Differentiating Exponential Functions From First Principles Key Questions How can I find the derivative of y = ex from first principles? 202 0 obj <> endobj Note that as x increases by one unit, from 3 to 2, the value of y decreases from 9 to 4. 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. NOTE: For a straight line: the rate of change of y with respect to x is the same as the gradient of the line. & = n2^{n-1}.\ _\square This is also known as the first derivative of the function. 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. ", and the Derivative Calculator will show the result below. MH-SET (Assistant Professor) Test Series 2021, CTET & State TET - Previous Year Papers (180+), All TGT Previous Year Paper Test Series (220+). Differentiation From First Principles This section looks at calculus and differentiation from first principles. example The x coordinate of Q is x + dx where dx is the symbol we use for a small change, or small increment in x. Differentiation from first principles involves using \(\frac{\Delta y}{\Delta x}\) to calculate the gradient of a function. 244 0 obj <>stream You will see that these final answers are the same as taking derivatives. It is also known as the delta method. The rate of change of y with respect to x is not a constant. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin . 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 . Abstract. + #, Differentiating Exponential Functions with Calculators, Differentiating Exponential Functions with Base e, Differentiating Exponential Functions with Other Bases. Example: The derivative of a displacement function is velocity. 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). This section looks at calculus and differentiation from first principles. It means that the slope of the tangent line is equal to the limit of the difference quotient as h approaches zero. Nie wieder prokastinieren mit unseren Lernerinnerungen. This is also referred to as the derivative of y with respect to x. 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. 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. For more about how to use the Derivative Calculator, go to "Help" or take a look at the examples. They are also useful to find Definite Integral by Parts, Exponential Function, Trigonometric Functions, etc. & = \lim_{h \to 0^+} \frac{ \sin (0 + h) - (0) }{h} \\ First principle of derivatives refers to using algebra to find a general expression for the slope of a curve. If we substitute the equations in the hint above, we get: \[\lim_{h\to 0} \frac{\cos x(\cos h - 1)}{h} - \frac{\sin x \cdot \sin h}{h} \rightarrow \lim_{h \to 0} \cos x (\frac{\cos h -1 }{h}) - \sin x (\frac{\sin h}{h}) \rightarrow \lim_{h \to 0} \cos x(0) - \sin x (1)\], \[\lim_{h \to 0} \cos x(0) - \sin x (1) = \lim_{h \to 0} (-\sin x)\]. Log in. As we let dx become zero we are left with just 2x, and this is the formula for the gradient of the tangent at P. We have a concise way of expressing the fact that we are letting dx approach zero. \[\displaystyle f'(1) =\lim_{h \to 0}\frac{f(1+h) - f(1)}{h} = p \ (\text{call it }p).\]. Our calculator allows you to check your solutions to calculus exercises. So even for a simple function like y = x2 we see that y is not changing constantly with x. Since there are no more h variables in the equation above, we can drop the \(\lim_{h \to 0}\), and with that we get the final equation of: Let's look at two examples, one easy and one a little more difficult. Please enable JavaScript. 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 \(a\). We want to measure the rate of change of a function \( y = f(x) \) with respect to its variable \( x \). Derivative by first principle refers to using algebra to find a general expression for the slope of a curve. For each calculated derivative, the LaTeX representations of the resulting mathematical expressions are tagged in the HTML code so that highlighting is possible. We denote derivatives as \({dy\over{dx}}\(\), which represents its very definition. We now have a formula that we can use to differentiate a function by first principles. Doing this requires using the angle sum formula for sin, as well as trigonometric limits. # " " = lim_{h to 0} {e^xe^h-e^(x)}/{h} # Use parentheses, if necessary, e.g. "a/(b+c)". & = \boxed{0}. endstream endobj startxref These changes are usually quite small, as Fig. While the first derivative can tell us if the function is increasing or decreasing, the second derivative. We know that, \(f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\). Suppose we choose point Q so that PR = 0.1. We take two points and calculate the change in y divided by the change in x. This book makes you realize that Calculus isn't that tough after all. Full curriculum of exercises and videos. The derivative is a powerful tool with many applications. Example : We shall perform the calculation for the curve y = x2 at the point, P, where x = 3. \(\begin{matrix} f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{f(-7+h)f(-7)\over{h}}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{|(-7+h)+7|-0\over{h}}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{|h|\over{h}}\\ \text{as h < 0 in this case}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{-h\over{h}}\\ f_{-}(-7)=-1\\ \text{On the other hand}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{f(-7+h)f(-7)\over{h}}\\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{|(-7+h)+7|-0\over{h}}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{|h|\over{h}}\\ \text{as h > 0 in this case}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{h\over{h}}\\ f_{+}(-7)=1\\ \therefore{f_{-}(a)\neq{f_{+}(a)}} \end{matrix}\), Therefore, f(x) it is not differentiable at x = 7, Learn about Derivative of Cos3x and Derivative of Root x. Create beautiful notes faster than ever before. You can try deriving those using the principle for further exercise to get acquainted with evaluating the derivative via the limit. Once you've done that, refresh this page to start using Wolfram|Alpha. It is also known as the delta method. There is a traditional method to differentiate functions, however, we will be concentrating on finding the gradient still through differentiation but from first principles. David Scherfgen 2023 all rights reserved. here we need to use some standard limits: \(\lim_{h \to 0} \frac{\sin h}{h} = 1\), and \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\). Sign up, Existing user? any help would be appreciated. For those with a technical background, the following section explains how the Derivative Calculator works. Differentiation from First Principles. This special exponential function with Euler's number, #e#, is the only function that remains unchanged when differentiated. 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) \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=tanx\\ f(x+h)=tan(x+h)\\ f(x+h)f(x)= tan(x+h) tan(x) = {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\\ {f(x+h) f(x)\over{h}}={ {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosxsin(x+h) sinxcos(x+h)\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {{sin(2x+h)+sinh\over{2}} {sin(2x+h)-sinh\over{2}}\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {1\over{cosxcos(x+h)}}\\ =1\times{1\over{cosx\times{cosx}}}\\ ={1\over{cos^2x}}\\ ={sec^2x}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = {sec^2x}\\ f(x)={dy\over{dx}} = {d(tanx)\over{dx}} = {sec^2x} \end{matrix}\), \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=sin5x\\ f(x+h)=sin(5x+5h)\\ f(x+h)f(x)= sin(5x+5h) sin(5x) = sin5xcos5h + cos5xsin5h sin5x\\ = sin5x(cos5h-1) + cos5xsin5h\\ {f(x+h) f(x)\over{h}}={ sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sin5x(cos5h-1)\over{h}} + \lim _{h{\rightarrow}0} {cos5xsin5h\over{h}}\\ = sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} + cos5x \lim _{h{\rightarrow}0} {sin5h\over{h}}\\ \text{Put h = 0 in first limit}\\ sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} = sin5x\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} 5\times{{d\over{dh}}sin5h\over{{d\over{dh}}5h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} {5cos5h\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \times5 = 5cos5x \end{matrix}\).