Differentiation of some elementary functions

PYQ-2023-MOD-Q1, PYQ-2023-MOD-Q6, PYQ-2023-Trigonometric_Equations-Q2

• $$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$$

• $$\frac{d}{d x}\left(a^{x}\right)=a^{x} \ell n a$$

• $$\frac{d}{d x}(\ell n|x|)=\frac{1}{x}$$

• $$\frac{d}{d x}\left(\log _{a} x\right)=\frac{1}{x \ell n}$$

• $$\frac{d}{d x}(\sin x)=\cos x$$

• $$\frac{d}{d x}(\cos x)=-\sin x $$

• $$ \frac{d}{d x}(\sec x)=\sec x \tan x$$

• $$\frac{d}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cot x$$

• $$\frac{d}{d x}(\tan x)=\sec ^{2} x$$

• $$\frac{d}{d x}(\cot x)=-\operatorname{cosec}^{2} x$$

Basic theorems

PYQ-2023-MOD-Q5

• $$\frac{d}{d x}(f \pm g)=f^{\prime}(x) \pm g^{\prime}(x)$$

• $$\frac{d}{d x}(k f(x))=k \frac{d}{d x} f(x)$$

• $$\frac{d}{d x}(f(x) \cdot g(x))=f(x) g^{\prime}(x)+g(x) f^{\prime}(x)$$

• $$\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)}$$

• $$\frac{d}{d x}(f(g(x)))=f^{\prime}(g(x)) g^{\prime}(x)$$

Derivative of inverse trigonometric functions

• $$ \frac{d \sin ^{-1} x}{d x}=\frac{1}{\sqrt{1-x^{2}}}, ~ ~ \frac{d \cos ^{-1} x}{d x}=-\frac{1}{\sqrt{1-x^{2}}}, for -1<x<1$$

$$ \frac{d \tan ^{-1} x}{d x}=\frac{1}{1+x^{2}},~ ~ \frac{d \cot ^{-1} x}{d x}=-\frac{1}{1+x^{2}} \quad(x \in R) $$

• $$ \frac{d \sec ^{-1} x}{d x}=\frac{1}{|x| \sqrt{x^{2}-1}},~ ~ \frac{d ~cosec^{-1} x}{d x}=-\frac{1}{|x| \sqrt{x^{2}-1}}, \text{for } ~~x \in(-\infty ,-1) \cup (1 , \infty )$$

Differentiation using substitution

$ \quad$ Following substitutions are normally used to sumplify these expression.

• $$\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}} ~ \text{by substituting}~ x=a \tan \theta, ~ \text{where~} -\frac{\pi}{2}<\theta<\frac{\pi}{2}$$

• $$\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} ~ \text{by substituting}~ x=a \sin \theta, ~ \text{where~} -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$

• $$\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}} \quad ~ \text{by substituting}~ \mathrm{x}=\mathrm{a} \sec \theta, ~ \text{where~} \theta \in[0, \pi], \quad \theta \neq \frac{\pi}{2}$$

• $$\sqrt{\frac{x+a}{a-x}} ~ \text{by substituting}~ x=a \cos \theta, ~ \text{where~} \theta \in(0, \pi].$$

Parametric differentiation

$\quad$ If $y=f(\theta)$ & $ x=g(\theta)$ where $\theta$ is a parameter, then $\frac{d y}{d x}=\frac{\frac{d y }{d \theta}}{\frac{d x }{d \theta} }$.

Derivative of one function with respect to another

$\quad$ Let $y=f(x) ; z=g(x)$ then $\frac{d y}{d z}=\frac{\frac{d y }{d x}}{\frac{d z }{d x}}=\frac{f^{\prime}(x)}{g^{\prime}(x)}$.

$\quad$ If $F(x)= \left|\begin{matrix} f(x) & g(x) & h(x) \\ l(x) & m(x) & n(x) \\ u(x) & v(x) & w(x) \end{matrix}\right|,~ $ where f, g, h, l, m, n, u, v, w are differentiable then,

$$F’(x)= \left|\begin{matrix} f’(x) & g’(x) & h’(x) \\ l(x) & m(x) & n(x) \\ u(x) & v(x) & w(x) \end{matrix}\right|+ \left|\begin{matrix} f(x) & g(x) & h(x) \\ l’(x) & m’(x) & n’(x) \\ u(x) & v(x) & w(x) \end{matrix}\right|+\left|\begin{matrix} f(x) & g(x) & h(x) \\ l(x) & m(x) & n(x) \\ u’(x) & v’(x) & w’(x) \end{matrix}\right| $$

$\quad$ The derivative of f(x) with respect to x is given by: $f^{\prime}(x)= \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Elementary differentiation:

• $$\frac{\mathrm{d}}{\mathrm{dx}} (\text{constant}) =0$$

• $$\frac{d}{d x}\left(x^n\right)=n x^{n-1}$$

• $$\frac{d}{d x}\left(e^x\right)=e^x$$

• $$\frac{d}{d x}\left(a^x\right)=a^x \log _e a$$

• $$\frac{\mathrm{d}}{\mathrm{dx}}\left(\log _{\mathrm{e}} \mathrm{x}\right)=\frac{1}{\mathrm{x}}$$

• $$\frac{d}{d x}\left(\log _a x\right)=\frac{1}{x \log _e a}$$

• $$\frac{d}{d x}(\sin x)=\cos x$$

• $$\frac{d}{d x}(\cos x)=-\sin x$$

• $$\frac{d}{d x}(\tan x)=\sec ^2 x$$

• $$\frac{d}{d x}(\cot x)=-\operatorname{cosec}^2 x$$

• $$\frac{d}{d x}(\sec x)=\sec x \tan x$$

• $$\frac{d}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cot x$$

• $$\frac{d}{d x}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1- \mathrm{x}^2}},-1<\mathrm{x}<1$$

• $$\frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}},-1<x<1$$

• $$\frac{d}{d x}\left(\tan ^{-1} \mathrm{x}\right)=\frac{1}{1+\mathrm{x}^2}, x \in R$$

• $$\frac{\mathrm{d}}{\mathrm{dx}}\left(\cot ^{-1} \mathrm{x}\right)=-\frac{1}{1+\mathrm{x}^2}, \forall \mathrm{x} \in \mathrm{R}$$

• $$\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}}|x|>1$$

• $$\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2-1}}|x|>1$$

• $$\frac{d}{d x}(\sinh x)=\cosh x$$

• $$\frac{d}{d x}(\cosh x)=\sinh x$$

• $$\frac{d}{d x}(\tanh x)=\operatorname{sech}^2 x$$

• $$\frac{d}{d x}(\operatorname{coth} x)=-\operatorname{cosech}^2 x$$

• $$\frac{d}{d x}(\sinh x)=\cosh x$$

• $$\frac{d}{d x}(\cosh x)=\sinh x$$

• $$\frac{d}{d x}(\tanh x)=\operatorname{sech}^2 x$$

• $$\frac{d}{d x}(\operatorname{coth} x)=-\operatorname{cosech}^2 x$$

• $$\frac{d}{d x}(\operatorname{sech} x)=-\operatorname{sech} x \tanh x$$

• $$\frac{d}{d x}(\operatorname{cosech} x)=-\operatorname{cosech} x \operatorname{coth} x$$

• $$\frac{d}{d x}\left(\sinh ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1+\mathrm{x}^2}}, \forall \mathrm{x} \in \mathrm{R}$$

• $$\frac{d}{d x}\left(\cosh ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{\mathrm{x}^2-1}},|\mathrm{x}|>1$$

• $$\frac{d}{d x}\left(\tanh ^{-1} \mathrm{x}\right)=\frac{1}{1-\mathrm{x}^2}, \mathrm{x} \pm 1$$

• $$\frac{d}{d x}\left(\operatorname{coth}^{-1} x\right)=\frac{1}{x^2-1}, x \neq \pm 1$$

• $$\frac{d}{d x}\left(\operatorname{sech}^{-1} x\right)=-\frac{1}{|x| \sqrt{1-x^2}},|x|<1$$

• $$\frac{d}{d x}\left(\operatorname{cosech}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2+1}}, \forall x \in R$$

• $$\frac{d}{d x}\left(e^{a x} \sin b x\right)=e^{a x}(a \sin b x+b \cos b x)=\sqrt{a^2+b^2} e^{a x} \sin \left(b x+\tan ^{-1} b / a\right)$$

• $$\frac{d}{d x}\left(e^{a x} \cos b x\right)=e^{a x}(a \cos b x-b \sin b x)=\sqrt{a^2+b^2} e^{a x} \cos \left(b x+\tan ^{-1} b / a\right)$$

• $$\frac{d}{d x}|x|=\frac{x}{|x|}(x \neq 0)$$

• $$\frac{d}{d x} \log |x|=\frac{1}{x},(x \neq 0)$$

• $$\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x}]=0, \forall \mathrm{x} \in-\mathrm{I}$$ (where [ . ] denotes greatest integer function)

• $$\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}=1, \forall \mathrm{x} \in \mathrm{R} \text{~} $$ { . } where denotes fractional part function

Product rule:

PYQ-2023-MOD-Q6

$\quad$ $$(gh)^{\prime} = gh^{\prime} + hg^{\prime} $$

Some useful substitutions in finding derivatives:

$$ \text{Function} \quad \rightarrow \quad \text{Substitution} $$

• $$\sqrt{a^2-x^2}\quad \rightarrow \quad x=a \sin \theta \text{~ or ~ } a \cos \theta $$

• $$\sqrt{x^2+a^2}\quad \rightarrow \quad x=a \tan \theta \text{~ or ~ } a \cot \theta $$

• $$\sqrt{x^2-a^2}\quad \rightarrow \quad x=a \sec \theta \text{~ or ~ } a \operatorname{cosec} \theta$$

• $$\sqrt{\frac{a-x}{a+x}}\quad \quad \rightarrow \quad x=a \cos 2 \theta$$

• $$\sqrt{\frac{a^2-x^2}{a^2+x^2}}\quad \quad \rightarrow \quad x^2=a^2 \cos 2 \theta$$

• $$\sqrt{a x-x^2}\quad \rightarrow \quad x=a \sin ^2 \theta$$

• $$\sqrt{\frac{x}{a+x}}\quad \qquad \rightarrow \quad x=a \tan ^2 \theta$$

• $$\sqrt{\frac{x}{a-x}}\quad \quad \quad \rightarrow \quad x=a \sin ^2 \theta$$

• $$\sqrt{(x-a)(x-b)}\quad \rightarrow \quad x=a \sec ^2 \theta-b \tan ^2 \theta$$

• $$\sqrt{(x-a)(b-x)}\quad \rightarrow \quad x=a \cos ^2 \theta+b \sin ^2 \theta$$

Successive differentiation:

PYQ-2023-MOD-Q4, PYQ-2023-MOD-Q6, PYQ-2023-AOD-Q11

$\quad$ If the first derivative $\frac{dy}{dx}$ of a function y = f(x) is also a differentiable function, then it can be further differentiated with respect to x.

$\quad$ $\frac{d^n y}{dx^n}$ denotes the n-th derivative of y.

Logarithmic differentiation:

$\quad$ $y = [f(x)]^{g(x)}$ then, taking logarithm, we get $\log y = g(x) \log f(x)$ then, $y^{\prime} = y (\frac{g}{f} f^{\prime} + g^{\prime} \log(f))$

Differentiation of implicit function:

PYQ-2023-MOD-Q2

$\quad$ If in an equation, both x and y occur together, i.e. f(x, y) = 0, and the equation cannot be solved for either x or y, then x (or y) is called the implicit function of y (or x).

• Rule for finding the derivative:-

  (i) Every term of f(x, y) = 0 should be differentiated with respect to x.

  (ii) The value of $\frac{dy}{dx}$ should be obtained by rearranging the terms