Question

The derivative of ${\sin ^3}x{\cos ^3}x$ is

Answer 1

Hint : As we can see that the above question consists of trigonometric ratios as sine and cosine are trigonometric ratios. We have to find the derivative of the given expression with the help of some trigonometric identities such as $\sin 2x = 2\sin x\cos x$ . We will multiply the numerator and denominator with suitable terms to solve it further.

Complete step-by-step answer :
Here we have to solve ${\sin ^3}x{\cos ^3}x$ . We know the trigonometric identity that
$\sin 2x = 2\sin x\cos x$
So we will multiply the denominator and numerator with ${2^3}$ and we have
$\dfrac{d}{{dx}}\left( {\dfrac{{{2^3}{{\sin }^3}{{\cos }^3}}}{{{2^3}}}} \right) \Rightarrow \dfrac{1}{8}\dfrac{d}{{dx}}({2^3}{\sin ^3}{\cos ^3})$ .
Using the help of trigonometric identity we can write the term as
$\dfrac{1}{8}\dfrac{d}{{dx}}{\left( {\sin 2x} \right)^3}$ . We will now first solve the second part of the expression i.e. $\dfrac{d}{{dx}}{\left( {\sin 2x} \right)^3}$ .
Now we can differentiate with the help of chain rule $\dfrac{d}{{dx}}[f(g(x))]$ is
$f'(g(x))g'(x)$ .
By applying this we can write
$3{\sin ^2}(2x)\left( {\cos ({2x})} \right)\dfrac{d}{{dx}}(2x) \Rightarrow 3{\sin ^2}(2x)\left( {\cos (2x)} \right)\dfrac{d}{{dx}}(2x)$ .
Since we know that $2$ is a constant with respect to $x$ , so the derivative of $2x$ is $2\dfrac{d}{{dx}}(x)$ .
We can put this in the expression and we have
$3{\sin ^2}(2x)\left( {\cos (2x)} \right)2\dfrac{d}{{dx}}(x)$ .
On further solving we can write
$2 \times 3{\sin ^2}(2x)\cos (2x)\dfrac{d}{{dx}}(x)$ .
We can now further differentiate it using the power rule which states that
$\dfrac{d}{{dx}}({x^n})$ is $n{x^{n - 1}}$ , here $n = 1$ . So it gives us value
$6{\sin ^2}(2x)\cos 2x \times 1$ .
By putting this back in the original equation we have
$\dfrac{1}{8} \times 6{\sin ^2}(2x)\cos 2x = \dfrac{3}{4}{\sin ^2}(2x)\cos 2x$.
Hence the required answer is $\dfrac{3}{4}{\sin ^2}(2x)\cos 2x$.

Note : We should note in the chain rule $\dfrac{d}{{dx}}[f(g(x))]$ is $f'(g(x))g'(x)$ , we have assumed $f(x) = \sin (x)$ and $g(x) = 2x$ . We have also used the formula i.e. $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ . When we solve $\dfrac{d}{{du}}(\sin ({u_2})) = \cos \left( {{u_2}} \right)$ , as we have assumed that ${u_2}$ as $2x$ . Before solving this kind of question we should have the clear concept of derivatives and their formulas.