Question

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

Answer 1

$\sin \left( {\cos x} \right)$ 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 $- \sin x\cos \left( {\cos x} \right)$.