Question

What is the derivative of ${\sin ^2}(x)?$

Answer 1

In this question we need to find the derivative of the given trigonometric function. We will use the chain rule to solve this question. We should know the derivative of the function i.e. $\dfrac{d}{{dx}}{u^2} = 2u$. So here we will assume that $u = \sin (x)$ and then we will differentiate this.

Complete step by step answer:
We will use the chain rule to solve this question. We know that the chain rule is useful for finding the derivatives of a function which could have been differentiated if it was $x$.The formula of chain rule is:
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
where, $\dfrac{{dy}}{{du}}$ is the derivative of the inner function and $\dfrac{{du}}{{dx}}$ is the derivative of the outer function.
So we will use some of the basic differentiation rules to solve this. We know that the derivative of sine is cosine i.e.
$\dfrac{d}{{dx}}\sin (x) = \cos (x)$
Also we know how to differentiate ${x^2}$.

We use the power rule of differentiation to solve this. It says that:
$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
So here we have
$n = 2$, by putting the values in the formula we have:
$2{x^{2 - 1}} = 2x$
Hence the derivative of ${x^2}$ is $2x$.
Here let us assume that
$u = \sin (x)$
We will find the derivative of the outer function.

So we can write that
$\dfrac{d}{{dx}}{u^2} = 2u$
Now by putting the value, we can write
$\dfrac{d}{{dx}}{\left( {\sin (x)} \right)^2} = 2\sin x$
As we have solved this above by using the product rule.
Now we know the derivative of the inner function i.e. $\sin x$
$\dfrac{d}{{dx}}\sin (x) = \cos (x)$
Now we can combine both the steps through multiplication to get the derivative i.e.
$2u\cos (x) = 2\sin (x)\cos (x)$

Hence we have $\dfrac{d}{{dx}}{\sin ^2}(x) = 2\sin (x)\cos (x)$.

Note: We can also solve the above question by using the product rule of two different functions. The product rule for differentiation states that the derivative of $f(x) \cdot g(x)$ is $f'(x)g(x) + f(x)g'(x)$ .
First let us assume that
$F(x) = {\sin ^2}x$
We should note that we can write
${\sin ^2}(x) = \sin (x) \times \sin (x)$
So we can set $f(x)$ and $g(x)$ as $\sin (x)$, it means that,
$F(x) = f(x)g(x)$, we will now apply the product rule.
By putting the values in the formula we can write
$f'(x)\sin (x) + \sin (x)g'(x)$
We should know that
$f'(x) = g'(x)$ and the derivative of
$\dfrac{d}{{dx}}\sin x = \cos x$
So by putting this all together we can write
$f'(x) = g'(x) = \cos (x)$
By substituting the value in the above expression we can write:
$\cos (x)\sin (x) + \sin (x)\cos (x)$
It can be written as:
$2\sin (x)\cos (x)$
Hence we also get the same answer by using the product rule of differentiation.