Question

pls give me step by step explanation and solve correctly ​

Answer 1

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

Given that,

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

and

$\red{\rm :\longmapsto\:f'(x) = \dfrac{1 + x}{1 - x}}$

Now,

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

Let assume that

$\red{\rm :\longmapsto\: {sin}^{2}x = z}$

So,

$\rm :\longmapsto\:y = f(z)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}f(z)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = f'(z)\dfrac{dz}{dx}$

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

$\red{\bigg \{ \because \:f'(x) = \dfrac{1 + x}{1 - x} \: \: \: and \: z \: = \: {sin}^{2}x \bigg \}}$

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

We know,

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1 + {sin}^{2} x}{{cos}^{2}x} \:2 \: sinx \dfrac{d}{dx}sinx$

We know,

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

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1 + {sin}^{2} x}{{cos}^{2}x} \:2 \: sinx \: cosx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1 + {sin}^{2} x}{{cos}^{2}x} \: sin2x$

can be rewritten as

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{(1 + {sin}^{2} x) \: sin2x}{{cos}^{2}x} \:$

Additional Information :-

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

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

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

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

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

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

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

Answer 2

Answer:

give that,

$y = f( \sin {}^{2} x)$

and

$f(x) = \frac{1 + x}{1 + x}$

now,

$y = f (sin {}^{2} x)$

let assume that

$\sin {}^{2} x = z$

so,

$y = f(z)$

on differentiating both sides W. r. t. x, we get

$\frac{d}{dx} y = \frac{d}{dx} f(z)$

$\frac{dy}{dx} = f(z) \frac{dz}{dx}$

$\frac{dy}{dx} = \frac{1 + z$