Question

Find the derivative of ${\left( {\sin x} \right)^n}$.

Hint-Here, we will proceed by differentiating the given function with respect to x and then we will use some formulas of differentiation which are $\dfrac{d}{{dx}}\left[ {{{\left( {f(x)} \right)}^n}} \right] = \left[ {n{{\left( {f(x)} \right)}^{n - 1}}} \right]\left[ {\dfrac{d}{{dx}}\left( {f(x)} \right)} \right]$ and $\dfrac{d}{{dx}}\left[ {\sin x} \right] = \cos x$.

Let us suppose the given function whose derivative is required as $y = {\left( {\sin x} \right)^n}$
Differentiating the above equation on both sides with respect to x, we get
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {{{\left( {\sin x} \right)}^n}} \right]{\text{ }} \to (1)$
As we know that $\dfrac{d}{{dx}}\left[ {{{\left( {f(x)} \right)}^n}} \right] = \left[ {n{{\left( {f(x)} \right)}^{n - 1}}} \right]\left[ {\dfrac{d}{{dx}}\left( {f(x)} \right)} \right]$
Using the above formula, equation (1) becomes
$\dfrac{{dy}}{{dx}} = n{\left( {\sin x} \right)^{n - 1}}\dfrac{d}{{dx}}\left[ {\sin x} \right]{\text{ }} \to {\text{(2)}}$
Also we know that $\dfrac{d}{{dx}}\left[ {\sin x} \right] = \cos x$
Using the above formula, equation (2) becomes
$\dfrac{{dy}}{{dx}} = n{\left( {\sin x} \right)^{n - 1}}\left[ {\cos x} \right]$
Hence, the derivative of the function ${\left( {\sin x} \right)^n}$ is $n{\left( {\sin x} \right)^{n - 1}}\left[ {\cos x} \right]$.

Note- In this particular problem, the function whose derivative is required consists of a constant i.e., n and the variable is x. Here, the derivative means the first derivative of the given function in x. If we would have been asked for the second derivative then the first derivative obtained i.e., $\dfrac{{dy}}{{dx}} = n{\left( {\sin x} \right)^{n - 1}}\left[ {\cos x} \right]$ needed to be differentiated once again with respect to x.