Differentiation of Functions

Differentiation is a method to find rates of change. It is an important topic for the JEE exam. Derivative of a function y = f(x) of a variable x is the rate of change of y with respect to rate of change of x. This article helps you to learn the derivative of a function, standard derivatives, theorems of derivatives, differentiation of implicit function and higher order derivatives along with solved examples. 

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How to Differentiate a Function

The differentiation of a function is a way to show the rate of change of a function at a given point. For real-valued functions, it is the slope of the tangent line at a point on a graph.

The derivative of y with respect to x is defined as the change in y over the change in x, as the distance between x0 and x1{\displaystyle x_{0}}\ and \ {\displaystyle x_{1}} becomes infinitely small (infinitesimal).  The derivative is often written as dydx{\displaystyle {\frac {dy}{dx}}}.

In mathematical terms,

f(a)=limh0f(a+h)f(a)h{\displaystyle f'(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}}

That is, as the distance between the two x points (h) becomes closer to zero, the slope of the line between them comes closer to resembling a tangent line.

Standard Derivatives

1] dx(un)=nun1dudx{dx}\left(u^n\right)=nu^{n-1}\frac{du}{dx}

2] ddx(c)=0,  where c is a constant\frac{d}{dx}\left(c\right)=0,\; \text{where c is a constant}

3] If y=F(u) where u=f(x) then dydx=dydu×dudx\text{If } y=F(u)\text{ where }u=f(x)\text{ then }\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}

(Chain Rule or Function of a Function Rule)

4] Derivatives of trigonometric functions

ddx(sinu)=cosududxddx(cosu)=sinududxddx(tanu)=sec2ududxddx(secu)=secutanududxddx(cosec u)=cosecu cotu dudxddx(cot u)=cosec2u dudx\frac{d}{dx}\left(\sin u\right)=\cos u\frac{du}{dx}\\ \frac{d}{dx}\left(\cos u\right)=-\sin u\frac{du}{dx}\\ \frac{d}{dx}\left(\tan u\right)=\sec^2 u\frac{du}{dx}\\ \frac{d}{dx}\left(\sec u\right)=\sec u\tan u\frac{du}{dx}\\ \frac{d}{dx}(cosec\ u)=-cosecu\ cotu\ \frac{du}{dx}\\ \frac{d}{dx}(cot\ u)=-cosec^2u\ \frac{du}{dx}\\

5] Derivatives of inverse trigonometric functions

ddx(sin1u)=11u2dudx,1<u<1ddx(cos1u)=11u2dudx,1<u<1ddx(tan1u)=11+u2dudxddx cosec1u=1uu21dudxu>1ddx(sec1u)=1uu21dudxu>1ddx(cot1u)=11+u2dudx\frac{d}{dx}\left(\sin^{-1} u\right)=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -1<u<1\\ \frac{d}{dx}\left(\cos^{-1} u\right)=\frac{-1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -1<u<1\\ \frac{d}{dx}\left(\tan^{-1} u\right)=\frac{1}{1+u^2}\frac{du}{dx}\\ \frac{d}{dx}\ cosec^{-1}u=-\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u|>1\\ \frac{d}{dx}\left(\sec^{-1} u\right)=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u|>1\\ \frac{d}{dx}\left(\cot^{-1} u\right)=-\frac{1}{1+u^2}\frac{du}{dx}\\

6] Exponential and logarithmic functions
ddx(eu)=eududxddx(au)=aulnadudx,where a>0,a1ddx(lnu)=1ududxddx(lnau)=1ulnadudx, where a>0,a1\frac{d}{dx}\left(e^u\right)=e^u\frac{du}{dx}\\ \frac{d}{dx}\left(a^u\right)=a^u\ln a\frac{du}{dx}, \text{where }a>0, a\ne1\\ \frac{d}{dx}\left(\ln u\right)=\frac{1}{u}\frac{du}{dx}\\ \frac{d}{dx}\left(\ln_a u\right)=\frac{1}{u\ln a}\frac{du}{dx},\text{ where }a>0, a\ne 1\\

7] Hyperbolic functions

ddx(sinhu)=coshududxddx(coshu)=sinhududxddx(tanhu)=sech2ududxddx (sechu)=sechu tanhududxddx (cosechu)=cosechu cothududxddx (cothu)=cosech2ududx\frac{d}{dx}\left(\sinh u\right)=\cosh u\frac{du}{dx}\\ \frac{d}{dx}\left(\cosh u\right)=\sinh u\frac{du}{dx}\\ \frac{d}{dx}\left(\tanh u\right)=-sech^{2}u\frac{du}{dx}\\ \frac{d}{dx}\ (sech u)=-sechu\ tanhu\frac{du}{dx}\\ \frac{d}{dx}\ (cosech u)=-cosechu\ cothu\frac{du}{dx}\\ \frac{d}{dx}\ (coth u)=-cosech^{2}u\frac{du}{dx}\\

8] Inverse hyperbolic functions
ddxsinh1u=11+u2dudxddxcosh1u=1u21dudx,u>1ddxtanh1u=11u2dudx,u<1ddxcosech1u=1uu2+1dudxu0ddxsech1u=1u1u2dudx0<u<1ddxcoth1u=11u2dudx,u>1\frac{d}{dx}sinh^{-1} u=\frac{1}{\sqrt{1+u^2}}\frac{du}{dx}\\ \frac{d}{dx}cosh^{-1} u=\frac{1}{\sqrt{u^2-1}}\frac{du}{dx},\qquad u>1\\ \frac{d}{dx}tanh^{-1} u=\frac{1}{1-u^2}\frac{du}{dx},\qquad |u|<1\\ \frac{d}{dx}cosech^{-1} u=-\frac{1}{|u|\sqrt{u^2+1}}\frac{du}{dx}\qquad u\ne 0\\ \frac{d}{dx}sech^{-1} u=-\frac{1}{u\sqrt{1-u^2}}\frac{du}{dx}\qquad 0<u<1\\ \frac{d}{dx}coth^{-1} u=-\frac{1}{1-u^2}\frac{du}{dx},\qquad |u|>1\\

Some Standard Substitution

Expression Substitution

Simple tricks to solve complicated diffrential equations are listed here:

If function contains a2x2\sqrt{{{a}^{2}}-{{x}^{2}}} ; then substitute x = a sinθ or x = a cosθ

If function contains a2+x2\sqrt{{{a}^{2}}+{{x}^{2}}} ; then substitute x = a cotθ or x = a tan θ

If function contains x2+a2\sqrt{{{x}^{2}}+{{a}^{2}}} ; then substitute x = a cosecθ or x = a sec θ

Theorems of Derivatives

Find below some the important theorem results:

ddx[u±v]=dudx±dvdxddxuv=udvdx+vdudx(Product Rule)ddxuv=vdudxudvdxv2(Quotient Rule)ddx(kf(x))=kddx(f(x)),wherekisconstant\frac{d}{dx}[u\pm v]=\frac{du}{dx}\pm\frac{dv}{dx}\\ \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}\qquad\text{(Product Rule)}\\ \frac{d}{dx}\frac{u}{v}=\frac{v\frac{du}{dx}- u\frac{dv}{dx}}{v^2}\qquad\text{(Quotient Rule)}\\\frac{d}{dx}\left( k\,f(x) \right)=k\frac{d}{dx}(f(x)),\,\,where\,k\,is\,constant

 

Example 1: Find dydx\frac{dy}{dx}for y = x sinx log x

Solution: y=1×sinx×logx+xcosxlogx+xsinx×1x=sinxlogx+xcosxlogx+sinx{{y}^{‘}}=1\times \sin x\times \log x+x\cos x\log x+x\sin x\times \frac{1}{x}=\sin x\,\log x+x\cos x\log x+\sin x

Example 2: Find dydxfory=sin(x2+1)\frac{dy}{dx}\,\,for\,y=\sin ({{x}^{2}}+1)

Solution: y=cos(x2+1)×2x{{y}^{‘}}=\cos ({{x}^{2}}+1)\times 2x

Differentiation of Implict Function

In implicit differentiation, also known as Chain Rule, differentiate both sides of an equation with two given variables by considering one of the as a function of the second variable. In short, differentiate the given function with respect to x and solve for dy/dx. Let us have a look on some the examples.

Example 1: If x2 + 2xy + y3 = 4, find dydx\frac{dy}{dx}

Solution: Differentiating both sides w.r.t. x, we get ddx(x2)+2ddx(xy)+ddx(y3)=ddx(4)\frac{d}{dx}({{x}^{2}})+2\frac{d}{dx}(xy)+\frac{d}{dx}({{y}^{3}})=\frac{d}{dx}(4) =2x+2xdydx+2y+3y2.dydx=0=2x+2x\frac{dy}{dx}+2y+3{{y}^{2}}.\frac{dy}{dx}=0 dydx=2(x+y)(2x+3y2)\Rightarrow \frac{dy}{dx}=\frac{-2(x+y)}{(2x+3{{y}^{2}})}

Example 2: Differentiable log sin x w.r.t cosx\sqrt{\cos x}

Solution: Let u = log sin x  and v = cosx\sqrt{\cos x} dudx=cotx&dvdx=sinx2cos\frac{du}{dx}=\cot x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\And \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{dv}{dx}=\frac{-\sin x}{2\sqrt{\cos }} dudx=du/dxdv/dx=cotxsinx2cosx=2coscotxcosec(x)\frac{du}{dx}=\frac{du/dx}{dv/dx}=\frac{\cot x}{\frac{-\sin x}{2\sqrt{\cos x}}}=-2\sqrt{\cos }\cot x\cos ec(x)

Higher order Derivatives

Differentiation process can be continued upto nth derivation of a function. Usually we deal with first order and second order derivatives of the functions.

dydx\frac{dy}{dx} is the first derivative of y w.r.t x

d2ydx2\frac{d^2y}{dx^2} is the second derivative of y w.r.t x

Similarly, finding the third, fourth, fifth and successive derivatives of any function, say g(x), which are known as higher-order derivatives of g(x).
nth order derivative numerical notation is g^n(x) or dnydxn\frac{d^ny}{dx^n}

Example: If y=etan1x,thenprovethat(1+x2)d2ydx2=(12x)dydxy={{e}^{{{\tan }^{-1}}x}},then\,prove\,that\,(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=(1-2x)\frac{dy}{dx}

Solution: y=etan1xdydx=dtan1x1+x2y={{e}^{{{\tan }^{-1}}x}}\Rightarrow \frac{dy}{dx}=\frac{{{d}^{{{\tan }^{-1}}}}x}{1+{{x}^{2}}} d2ydx2=(12x)etan1x(1+x2)2\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(1-2x){{e}^{ta{{n}^{-1}}x}}}{{{(1+{{x}^{2}})}^{2}}}

Again,

d2ydx2=(1+x2)=(12x)dydx\frac{{{d}^{2}}y}{d{{x}^{2}}}=(1+{{x}^{2}})=(1-2x)\frac{dy}{dx}

 

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