Question

Differentiate the given function with respect to $x$
$\sin \left( {{x}^{2}}+5 \right)$

Answer 1

Hint: To solve these types of questions where one function is inside the other function, then we have to do the differentiation of these functions with the help of chain rule. Chain rule’s general form for finding the derivative of complex function like $f\left( x \right)$ is
$\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=\dfrac{d}{dx}\left[ f\left( u \right) \right]\times \dfrac{du}{dx}$ where $u=g\left( x \right)$

Complete step-by-step answer:
Before proceeding to differentiate the function, we must know what is chain rule in differentiation. The chain rule is basically a formula to compute the derivative of a composite function. The general form of a composite function is $f\left( g\left( x \right) \right)$ .To differentiate such functions, we will use chain rule.
$\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=\dfrac{d}{du}\left[ f\left( u \right) \right]\times \dfrac{du}{dx}$ where
Since the function we have to differentiate is also a composite function, we can use chain rule here.

Let $f\left( x \right)=\sin \left( {{x}^{2}}+5 \right)$ . Now we have to differentiate this function with respect to $x$ . To do this, we will use the chain rule of differentiation.
Let the inner function, ${{x}^{2}}+5=4$ . Now the function becomes a variable in u i.e. we get $f\left( u \right)=\sin u$
$f\left( u \right)=\sin u................\left( i \right)$
Now, we will differentiate both sides of the equation with respect to $x$ . After differentiating, we get
$\dfrac{df\left( u \right)}{dx}=\dfrac{d\left( \sin u \right)}{dx}................\left( ii \right)$
Now, multiply both sides of the equation with $\dfrac{du}{du}$ .After multiplication, we will get:
$\dfrac{df\left( u \right)}{dx}\times \dfrac{du}{du}=\dfrac{d\left( \sin u \right)}{dx}\times \dfrac{du}{du}$
Now, we will rearrange these terms. After rearranging we get:
$\dfrac{df\left( u \right)}{du}\times \dfrac{du}{dx}=\dfrac{d\left( \sin u \right)}{du}\times \dfrac{du}{dx}..............\left( iii \right)$
Thus, we have got two terms of RHS. After finding the value of these terms, we will get the answer.
Calculation of $\dfrac{d\left( \sin u \right)}{du}$ : We have to calculate the differentiation of $\sin u$ with respect to u. Its value is equal to $\cos u$ .Therefore,
$\dfrac{d}{du}\left( \sin u \right)=\cos u..............\left( iv \right)$
Calculation of $\dfrac{du}{dx}$ : We know that $u={{x}^{2}}+5$ . We will differentiate both sides with respect to $x$ . After differentiation we get:
$\dfrac{du}{dx}=2u...............\left( v \right)$
Now, we will put the values of (iv) and (v) in (iii).
After putting we get:
$\dfrac{df\left( u \right)}{du}\times \dfrac{du}{dx}=2x\cos u$
Now, we will put the value of $u={{x}^{2}}+5$ on both sides.
After doing this we will get:

& \dfrac{d\sin \left( {{x}^{2}}+5 \right)}{{d\left( {{x}^{2}}+5 \right)}}\times \dfrac{ {d\left( {{x}^{2}}+5 \right)}}{dx}=2x\cos \left( {{x}^{2}}+5 \right) \\\ & \Rightarrow \dfrac{d}{dx}\left( \sin \left( {{x}^{2}}+5 \right) \right)=2x\cos \left( {{x}^{2}}+5 \right) \\\ \end{aligned}$$ This is our required answer. Note: We cannot use chain rule in every function. The necessary condition for the chain rule to be applied in the differentiation of $f\left( g\left( x \right) \right)$ is that both $f\left( x \right)$ and $g\left( x \right)$ should be differentiable. In our case both the sine function and ${{x}^{2}}+5$ is differentiable. That is why we can use chain rule.