Question

What is the derivative of \[\sin \left( {\cos x} \right)\]?

Answer 1

Hint:\[\sin \left( {\cos x} \right)\] is a composite function. We will use the concept of chain rule of differentiation to find the derivative of the given composite function. From the chain rule of differentiation, we know that \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = {f^1}\left( {g\left( x \right)} \right){g^1}\left( x \right)\]. Here, \[g\left( x \right)\] is \[\cos x\] and \[f\left( {g\left( x \right)} \right)\] is \[\sin \left( {\cos x} \right)\].Using this we will find the derivative of \[\sin \left( {\cos x} \right)\].

Complete step by step answer:
Given is a sine function in the form of \[\sin \left( {\cos x} \right)\].To find the derivative of \[\sin \left( {\cos x} \right)\], we will use the concept of chain rule of differentiation.From the chain rule of differentiation, we know that,
\[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = {f^1}\left( {g\left( x \right)} \right){g^1}\left( x \right)\].

As we know, the differentiation of \[\sin x\] is \[\cos x\].
Therefore, on differentiation of the first term of \[\sin \left( {\cos x} \right)\], we get \[\cos \left( {\cos x} \right)\].
Let us assume \[\cos \left( {\cos x} \right) = A\]. Now, as we know, the differentiation of \[\cos x\] is \[\left( { - \sin x} \right)\].
On differentiating the second function, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Therefore, we get differentiation of the second function as \[\left( { - \sin x} \right)\].

Let us assume \[\left( { - \sin x} \right) = B\]. We know that we have to multiply the result of both the differentiation to get the result i.e.,
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin \left( {\cos x} \right)} \right) = A \times B\]
Substituting the values of \[A\] and \[B\], we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin \left( {\cos x} \right)} \right) = \cos \left( {\cos x} \right) \times \left( { - \sin x} \right)\]
On rewriting we get,
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin \left( {\cos x} \right)} \right) = - \sin x\cos \left( {\cos x} \right)\]

Therefore, the derivative of \[\sin \left( {\cos x} \right)\] is \[ - \sin x\cos \left( {\cos x} \right)\].

Note:We can also solve this problem by taking \[\sin \left( {\cos x} \right)\] as \[y\] and then applying \[{\sin ^{ - 1}}\] on both the sides and then differentiating to find differentiation of \[y\] with respect to \[x\] i.e., \[\dfrac{{dy}}{{dx}}\].
Let \[y = \sin \left( {\cos x} \right)\].
Taking \[{\sin ^{ - 1}}\] on both the sides, we get
\[ \Rightarrow {\sin ^{ - 1}}y = \cos x\]

On differentiating both the side with respect to \[x\], we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}y} \right) = \dfrac{d}{{dx}}\left( {\cos x} \right) - - - (1)\]
As we know that \[\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\].
So, differentiation of \[{\sin ^{ - 1}}y\] with respect to \[x\] is given by,
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}y} \right) = \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}}\]
Also, \[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\].

Putting these values in \[(1)\], we get
\[ \Rightarrow \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}} = - \sin x\]
Putting the value of \[y\], we get
\[ \Rightarrow \dfrac{1}{{\sqrt {1 - {{\sin }^2}\left( {\cos x} \right)} }}\dfrac{{dy}}{{dx}} = - \sin x\]
As \[{\sin ^2}x + {\cos ^2}x = 1\], using this we get
\[ \Rightarrow \dfrac{1}{{\sqrt {{{\cos }^2}\left( {\cos x} \right)} }}\dfrac{{dy}}{{dx}} = - \sin x\]

On simplification,
\[ \Rightarrow \dfrac{1}{{\cos \left( {\cos x} \right)}}\dfrac{{dy}}{{dx}} = - \sin x\]
On rearranging, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin x\cos \left( {\cos x} \right)\]
Therefore, the derivative of \[\sin \left( {\cos x} \right)\] is \[ - \sin x\cos \left( {\cos x} \right)\].