Question

find the derivative (x + tan x)(x ^ 2 + 1)​

Answer 1

GIVEN :–

• A function → (x + tan x )( x² + 1 )

TO FIND :–

• Differentiate form = ?

SOLUTION :–

• Let the function –

$\\ \tt\implies P = (x + \tan x)( {x}^{2} + 1)$

$\\ \tt\implies P = {x}^{2} (x + \tan x) + 1(x + \tan x)$

$\\ \tt\implies P = {x}^{3} + {x}^{2} \tan x+ x + \tan x$

• Now Differentiate with respect to 'x' –

$\\ \tt\implies \dfrac{dP}{dx} = \dfrac{d}{dx}({x}^{3} + {x}^{2} \tan x+ x + \tan x)$

$\\ \tt\implies \dfrac{dP}{dx} = \dfrac{d({x}^{3})}{dx} + \dfrac{d({x}^{2} \tan x)}{dx} + \dfrac{d(x)}{dx} + \dfrac{d(\tan x)}{dx}$

• Using identity –

$\\ \large\longrightarrow \: { \boxed{ \tt \dfrac{d({x}^{n})}{dx} = n {x}^{n - 1}}}$

• And –

$\\ \large\longrightarrow \: { \boxed{ \tt \dfrac{d(u.v)}{dx} =u \dfrac{dv}{dx} + v \dfrac{du}{dx} }}$

• So that –

$\\ \tt\implies \dfrac{dP}{dx} = 3 {x}^{2} + \bigg \{ {x}^{2} \dfrac{d(\tan x)}{dx} + \tan x\dfrac{d( {x}^{2} )}{dx} \bigg \} +1+ \sec^{2}(x)$

$\\ \tt\implies \dfrac{dP}{dx} = 3 {x}^{2} + \bigg \{ {x}^{2}\sec^{2}(x)+ (\tan x)(2x)\bigg \} +1+ \sec^{2}(x)$

$\\ \tt\implies \dfrac{dP}{dx} = 3 {x}^{2} + {x}^{2}\sec^{2}(x)+2x\tan(x)+1+ \sec^{2}(x)$

$\\ \large\implies { \boxed{ \tt \dfrac{dP}{dx} = 3 {x}^{2} + ({x}^{2} + 1)\sec^{2}(x)+2x\tan(x)+1}}$