Differentiation

Table of Content

Standard Derivatives

1. $\frac{d({{x}^{n}})}{dx}=n{{x}^{n-1}},\,\,x\in R,\,\,\,n\in R,\,\,\,\,x>0$
2. $\frac{d}{dx}{{e}^{x}}={{e}^{x}}$
3. $\frac{d{{a}^{x}}}{dx}={{a}^{x}}\,\ell na$
4. $\frac{d}{dx}\ell nx=\frac{1}{x}$
5. d(sinx) = cosx
6. $\frac{d}{dx}(si{{n}^{-1}}x)=\frac{1}{\sqrt{1-{{x}^{2}}}}$.
7. $\frac{d}{dx}(co{{s}^{-1}}x)=\frac{-1}{\sqrt{1-{{x}^{2}}}}$

Some Standard Substitution

Expression Substitution

$\sqrt{{{a}^{2}}-{{x}^{2}}}$ $x=a\sin \theta \,\,or\,\,a\cos \theta$ $\sqrt{{{a}^{2}}+{{x}^{2}}}$ $x=a\tan \theta \,\,or\,\,a\cot \theta$ $\sqrt{{{x}^{2}}+{{a}^{2}}}$ $x=a\sec \theta \,\,or\,\,a\;cosec\theta$

If y = $y=\left( 1+{{x}^{{}^{1}/{}_{4}}} \right)\left( 1+{{x}^{{}^{1}/{}_{2}}} \right)\left( 1-{{x}^{{}^{1}/{}_{4}}} \right),$find $\frac{dy}{dx}$

Solution: $y=\left( 1+{{x}^{{}^{1}/{}_{4}}} \right)\left( 1-{{x}^{{}^{1}/{}_{4}}} \right)\left( 1+{{x}^{{}^{1}/{}_{2}}} \right)$ $=\left( 1+{{x}^{{}^{1}/{}_{2}}} \right)\left( 1+{{x}^{{}^{1}/{}_{2}}} \right)=1-x$ $\frac{dy}{dx}=-1$

Theorems of Derivatives

(a) d (f1 (x) + f2(x)} = $\frac{d}{dx}{{f}_{1}}(x)\pm \frac{d}{dx}{{f}_{2}}(x)$

(b) $\frac{d}{dx}\left( k\,f(x) \right)=k\frac{d}{dx}(f(x)),\,\,where\,k\,is\,constant$

(c) $\frac{d}{dx}\left( {{f}_{1}}(x)-f({{x}_{2}}) \right)={{f}_{1}}(x)\frac{d}{dx}{{f}_{2}}(x)+{{f}_{2}}(x)\frac{d}{dx}{{f}_{1}}(x)$

(d) $\frac{d}{dx}\left\{ \frac{{{f}_{1}}(x)}{{{f}_{2}}(x)} \right\}={{f}_{2}}(x)\,f_{1}^{1}\frac{(x)}{f_{2}^{2}(x)}={{f}_{1}}(x)f_{2}^{1}(x)$

Find $\frac{dy}{dx}$for y = x sinx log x

${{y}^{1}}=1\times \sin x\times \log x+x\cos x\log x+x\sin x\times \frac{1}{x}=\sin x\,\log x+x\cos x\log x+\sin x$

Find $\frac{dy}{dx}\,\,for\,y=\sin ({{x}^{2}}+1)$

${{y}^{1}}=\cos ({{x}^{2}}+1)\times 2x$

Differentiation of Implict Function

If x2 + 2xy + y3 = 4, find $\frac{dy}{dx}$

Differentiating both sides w.r.t. x, we get $\frac{d}{dx}({{x}^{2}})+2\frac{d}{dx}(xy)+\frac{d}{dx}({{y}^{3}})=\frac{d}{dx}(4)$ $=2x+2x\frac{dy}{dx}+2y+3{{y}^{2}}.\frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=\frac{-2(x+y)}{(2x+3{{y}^{2}})}$

Differentiable log sinx w.r.t $\sqrt{\cos x}$

Solution: Let u = log sinx , v = $sqrt{\cos x}$ $\frac{du}{dx}=\cot x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\And \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{dv}{dx}=\frac{-\sin x}{2\sqrt{\cos }}$ $\frac{du}{dx}=\frac{du/dx}{dv/dx}=\frac{\cot x}{\frac{-\sin x}{2\sqrt{\cos x}}}=-2\sqrt{\cos }\cot x\cos ec(x)$

Higher order Derivatives

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

If $y={{e}^{{{\tan }^{-1}}x}},then\,prove\,that\,(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=(1-2x)\frac{dy}{dx}$

Solution: $y={{e}^{{{\tan }^{-1}}x}}\Rightarrow \frac{dy}{dx}=\frac{{{d}^{{{\tan }^{-1}}}}x}{1+{{x}^{2}}}$ $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(1-2x){{e}^{ta{{n}^{-1}}x}}}{{{(1+{{x}^{2}})}^{2}}}$ $\frac{{{d}^{2}}y}{d{{x}^{2}}}=(1+{{x}^{2}})=(1-2x)\frac{dy}{dx}$

Practise This Question

π2014 cos2x+9 sin2xdx=