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
\(\begin{array}{l}{\displaystyle x_{0}}\ and \ {\displaystyle x_{1}}\end{array} \)
becomes infinitely small (infinitesimal). The derivative is often written as \(\begin{array}{l}{\displaystyle {\frac {dy}{dx}}}\end{array} \)
.
In mathematical terms,
If f is a real valued function and a is a point in its domain of definition. The derivative of f at a is defined by
\(\begin{array}{l}{\displaystyle f'(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}}\end{array} \)
Standard Derivatives
1]
\(\begin{array}{l}{dx}\left(u^n\right)=nu^{n-1}\frac{du}{dx}\end{array} \)
2]
\(\begin{array}{l}\frac{d}{dx}\left(c\right)=0,\; \text{where c is a constant}\end{array} \)
3]
\(\begin{array}{l}\text{If } y=F(u)\text{ where }u=f(x)\text{ then }\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\end{array} \)
(Chain Rule or Function of a Function Rule)
4] Derivatives of trigonometric functions
\(\begin{array}{l}\frac{d}{dx}\left(\sin u\right)=\cos u\frac{du}{dx}\\ \frac{d}{dx}\left(\cos u\right)=-\sin u\frac{du}{dx}\\ \frac{d}{dx}\left(\tan u\right)=\sec^2 u\frac{du}{dx}\\ \frac{d}{dx}\left(\sec u\right)=\sec u\tan u\frac{du}{dx}\\ \frac{d}{dx}(cosec\ u)=-cosecu\ cotu\ \frac{du}{dx}\\ \frac{d}{dx}(cot\ u)=-cosec^2u\ \frac{du}{dx}\\\end{array} \)
5] Derivatives of inverse trigonometric functions
\(\begin{array}{l}\frac{d}{dx}\left(\sin^{-1} u\right)=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -1<u<1\\ \frac{d}{dx}\left(\cos^{-1} u\right)=\frac{-1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -1<u<1\\ \frac{d}{dx}\left(\tan^{-1} u\right)=\frac{1}{1+u^2}\frac{du}{dx}\\ \frac{d}{dx}\ cosec^{-1}u=-\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u|>1\\ \frac{d}{dx}\left(\sec^{-1} u\right)=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u|>1\\ \frac{d}{dx}\left(\cot^{-1} u\right)=-\frac{1}{1+u^2}\frac{du}{dx}\\\end{array} \)
6] Exponential and logarithmic functions
\(\begin{array}{l}\frac{d}{dx}\left(e^u\right)=e^u\frac{du}{dx}\\ \frac{d}{dx}\left(a^u\right)=a^u\ln a\frac{du}{dx}, \text{where }a>0, a\ne1\\ \frac{d}{dx}\left(\ln u\right)=\frac{1}{u}\frac{du}{dx}\\ \frac{d}{dx}\left(\ln_a u\right)=\frac{1}{u\ln a}\frac{du}{dx},\text{ where }a>0, a\ne 1\\\end{array} \)
7] Hyperbolic functions
\(\begin{array}{l}\frac{d}{dx}\left(\sinh u\right)=\cosh u\frac{du}{dx}\\ \frac{d}{dx}\left(\cosh u\right)=\sinh u\frac{du}{dx}\\ \frac{d}{dx}\left(\tanh u\right)=sech^{2}u\frac{du}{dx}\\ \frac{d}{dx}\ (sech u)=-sechu\ tanhu\frac{du}{dx}\\ \frac{d}{dx}\ (cosech u)=-cosechu\ cothu\frac{du}{dx}\\ \frac{d}{dx}\ (coth u)=-cosech^{2}u\frac{du}{dx}\\\end{array} \)
8] Inverse hyperbolic functions
\(\begin{array}{l}\frac{d}{dx}sinh^{-1} u=\frac{1}{\sqrt{1+u^2}}\frac{du}{dx}\\ \frac{d}{dx}cosh^{-1} u=\frac{1}{\sqrt{u^2-1}}\frac{du}{dx},\qquad u>1\\ \frac{d}{dx}tanh^{-1} u=\frac{1}{1-u^2}\frac{du}{dx},\qquad |u|<1\\ \frac{d}{dx}cosech^{-1} u=-\frac{1}{|u|\sqrt{u^2+1}}\frac{du}{dx}\qquad u\ne 0\\ \frac{d}{dx}sech^{-1} u=-\frac{1}{u\sqrt{1-u^2}}\frac{du}{dx}\qquad 0<u<1\\ \frac{d}{dx}coth^{-1} u=\frac{1}{1-u^2}\frac{du}{dx},\qquad |u|>1\\\end{array} \)
Some Standard Substitution
Expression Substitution
Simple tricks to solve complicated differential equations are listed here:
If function contains
\(\begin{array}{l}\sqrt{{{a}^{2}}-{{x}^{2}}}\end{array} \)
; then substitute x = a sinθ or x = a cosθ
If function contains
\(\begin{array}{l}\sqrt{{{a}^{2}}+{{x}^{2}}}\end{array} \)
; then substitute x = a cotθ or x = a tan θ
If function contains
\(\begin{array}{l}\sqrt{{{x}^{2}}+{{a}^{2}}}\end{array} \)
; then substitute x = a cosecθ or x = a sec θ
Theorems of Derivatives
Find below some the important theorem results:
\(\begin{array}{l}\frac{d}{dx}[u\pm v]=\frac{du}{dx}\pm\frac{dv}{dx}\\ \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}\qquad\text{(Product Rule)}\\ \frac{d}{dx}\frac{u}{v}=\frac{v\frac{du}{dx}- u\frac{dv}{dx}}{v^2}\qquad\text{(Quotient Rule)}\\\frac{d}{dx}\left( k\,f(x) \right)=k\frac{d}{dx}(f(x)),\,\,where\,k\,is\,constant\end{array} \)
Example 1: Find \(\begin{array}{l}\frac{dy}{dx}\end{array} \)
for y = x sinx log x
Solution:
\(\begin{array}{l}{{y}^{‘}}=1\times \sin x\times \log x+x\cos x\log x+x\sin x\times \frac{1}{x}=\sin x\,\log x+x\cos x\log x+\sin x\end{array} \)
Example 2: Find \(\begin{array}{l}\frac{dy}{dx}\,\,for\,y=\sin ({{x}^{2}}+1)\end{array} \)
Solution:
\(\begin{array}{l}{{y}^{‘}}=\cos ({{x}^{2}}+1)\times 2x\end{array} \)
= 2x cos (x2 + 1)
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 variable 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
\(\begin{array}{l}\frac{dy}{dx}\end{array} \)
Solution: Differentiating both sides w.r.t. x, we get
\(\begin{array}{l}\frac{d}{dx}({{x}^{2}})+2\frac{d}{dx}(xy)+\frac{d}{dx}({{y}^{3}})=\frac{d}{dx}(4)\end{array} \)
\(\begin{array}{l}=2x+2x\frac{dy}{dx}+2y+3{{y}^{2}}.\frac{dy}{dx}=0\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=\frac{-2(x+y)}{(2x+3{{y}^{2}})}\end{array} \)
Example 2: Differentiable log sin x w.r.t
\(\begin{array}{l}\sqrt{\cos x}\end{array} \)
Solution: Let u = log sin x and v =
\(\begin{array}{l}\sqrt{\cos x}\end{array} \)
\(\begin{array}{l}\frac{du}{dx}=\cot x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\And \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{dv}{dx}=\frac{-\sin x}{2\sqrt{\cos x}}\end{array} \)
,
\(\begin{array}{l}\frac{du}{dx}=\frac{du/dx}{dv/dx}=\frac{\cot x}{\frac{-\sin x}{2\sqrt{\cos x}}}=-2\sqrt{\cos x}\cot x\cos ec(x)\end{array} \)
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.
\(\begin{array}{l}\frac{dy}{dx}\end{array} \)
is the first derivative of y w.r.t x
\(\begin{array}{l}\frac{d^2y}{dx^2}\end{array} \)
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 gn(x) or
\(\begin{array}{l}\frac{d^ny}{dx^n}\end{array} \)
Example: If
\(\begin{array}{l}y={{e}^{{{\tan }^{-1}}x}},then\,prove\,that\,(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=(1-2x)\frac{dy}{dx}\end{array} \)
Solution:
\(\begin{array}{l}y={{e}^{{{\tan }^{-1}}x}}\end{array} \)
dy/dx =
\(\begin{array}{l}{{e}^{{{\tan }^{-1}}x}}\frac{1}{1+x^{2}}\end{array} \)
=
\(\begin{array}{l}\frac{e^{tan^{-1}x}}{1+x^{2}}\end{array} \)
…(i)
d2y/dx2 =
\(\begin{array}{l}\frac{(1+x^{2})e^{tan^{-1}x}\frac{1}{1+x^{2}}-2xe^{tan^{-1}x}}{(1+x^{2})^{2}}\end{array} \)
=
\(\begin{array}{l}\frac{(1-2x)e^{tan^{-1}x}}{(1+x^{2})^{2}}\end{array} \)
(d2y/dx2 )(1 + x2) =
\(\begin{array}{l}\frac{(1-2x)e^{tan^{-1}x}}{(1+x^{2})}\end{array} \)
= (1-2x)dy/dx (from eqn (i))
Hence proved.
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