Question

$\huge\red{\boxed{\sf QuEsTioN}}$

Differentiate the function: cos (x²+1).​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:cos( {x}^{2} + 1)$

Let assume that

$\rm :\longmapsto\:y \: = \: cos( {x}^{2} + 1)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \: \dfrac{d}{dx} \: cos( {x}^{2} + 1)$

We know,

$\boxed{ \rm \:\dfrac{d}{dx}cosx \: = \: - \: sinx}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \: - \: sin( {x}^{2} + 1)\dfrac{d}{dx}( {x}^{2} + 1)$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \: - \: sin( {x}^{2} + 1)\bigg[\dfrac{d}{dx}{x}^{2} + \dfrac{d}{dx}1\bigg]$

We know,

$\boxed{ \rm \:\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}}$

and

$\boxed{ \rm \:\dfrac{d}{dx}k \: = \: 0}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \: - \: sin( {x}^{2} + 1)\bigg[ {2x}^{2 - 1} + 0\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \: - \: sin( {x}^{2} + 1)\bigg[ {2x}^{1}\bigg]$

$\bf\implies \:\dfrac{dy}{dx} \: = \: - \: 2x \: sin( {x}^{2} + 1)$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\huge\color{aqua}\ \boxed{\colorbox{black}{Answer : ♞ }}</p><p>.$

$\left(-\sin(x^{2}+1)\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+1)$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$\left(-\sin(x^{2}+1)\right)\times 2x^{2-1}$

SIMPLIFY:

$\left(-2x\right)\sin(x^{2}+1)$