Question

differentiate with respect to x using chain rule​

Answer 1

Answer:

2sin(2x+3) cos(2x+3)2

Answer 2

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

To differentiate

$\rm :\longmapsto\: {sin}^{2}(2x + 3)$

Let assume that,

$\rm :\longmapsto\:y = {sin}^{2}(2x + 3)$

can be rewritten as

$\rm :\longmapsto\:y = {\bigg(sin(2x + 3)\bigg) }^{2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} {\bigg(sin(2x + 3)\bigg) }^{2}$

We know,

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

So, using this,

$\rm :\longmapsto\:\dfrac{dy}{dx} =2 {\bigg(sin(2x + 3)\bigg) }^{2 - 1}\dfrac{d}{dx}sin(2x + 3)$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =2 {\bigg(sin(2x + 3)\bigg) }^{1}cos(2x + 3)\dfrac{d}{dx}(2x + 3)$

We know,

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$

and

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

$\rm :\longmapsto\:\dfrac{dy}{dx} =2 {sin(2x + 3) }cos(2x + 3)(2 \times 1 + 0)$

We know,

$\boxed{ \rm{ sin2x = 2sinx \: cosx}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = sin2(2x + 3) \times 2$

$\bf\implies \:\dfrac{dy}{dx} = 2sin(4x + 6)$

Additional Information :-

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

$\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = { - cosec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}secx = \: secx \: tanx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx} {e}^{x} = {e}^{x} }}$