Question

Find the derivative of the function with respect to x :

f(x) = Sin(Cos(x)²)


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Answer 1

Given:

f(x) = sin(cos(x²))

To Find:

Derivative of f(x) w.r.t. x

Solution:

We know that,

$\longrightarrow\rm{\dfrac{d}{dx}(f(x))=f'(x)}$

$\longrightarrow\rm{\dfrac{d}{dx}(sinx)=cosx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosx)=-sinx}$

$\longrightarrow\rm{\dfrac{d}{dx}(x^{n})=nx^{n-1}}$

$\rule{190}{1}$

On differentiating f(x) w.r.t. x, we get

$\longrightarrow\rm{f'(x)=\dfrac{d\Big(sin(cos(x^{2}))\Big)}{dx}}$

$\longrightarrow\rm{f'(x)=cos(cos(x^{2})).\dfrac{d\Big(cos(x^{2})\Big)}{dx}}$

$\rightarrow\rm{f'(x)=cos(cos(x^{2})).(-sin(x^{2})).\dfrac{d\big(x^{2}\big)}{dx}}$

$\longrightarrow\rm{f'(x)=cos(cos(x^{2})).(-sin(x^{2})).(2x)}$

$\longrightarrow\rm\green{f'(x)=-2x.cos(cos(x^{2})).sin(x^{2})}$

$\rule{190}{1}$

Formulae to Remember

$\longrightarrow\rm{\dfrac{d}{dx}(tanx)=sec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(cotx)=-cosec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(secx)=secx.tanx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosecx)=-cosecx.cotx}$

$\longrightarrow\rm{\dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(e^{x})=e^{x}}$

Answer 2

Simply use chain rule here

As we know that in chain rule ,we first diffrentiate the whole terms them we diffrentiate the number in bracket

So here

dy/dx {Sin(cos(x)^2)}=dy/dx sin(cos(x)^2)×dy/dx

(cos(x)^2)dy/dx=cos(cos(x)^2)×dy/dx (cos(x)^2)×dy/dx (x^2)

=Cos(cos(x)^2)×(-sin(x)^2)×2x

=-2xcos(cos(x)^2)(sin(x)^2)