Question

How do you find the derivative of the function $ {{\sin }^{3}}\left( 2x \right) $ ?

Answer 1

Hint:
We start solving the problem by assuming variables for the functions $ \sin 2x $ and $ 2x $ . We then recall the chain rule of differentiation as $ \dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{df}\times \dfrac{d\left( f\left( x \right) \right)}{dx} $ . We apply chain rule to the cube of the sine function followed by differentiating sine function and then differentiating the angle present inside the sine function. We then multiply all the obtained results of differentiation to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the derivative of the function $ {{\sin }^{3}}\left( 2x \right) $ .
Let us assume $ \sin 2x=z $ and $ 2x=y $ . Now, let us assume $ p={{\sin }^{3}}\left( 2x \right)\Leftrightarrow p={{z}^{3}} $ ---(1).
Now, let us differentiate both sides of equation (1) w.r.t x.
From the chain rule we know that $ \dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{df}\times \dfrac{d\left( f\left( x \right) \right)}{dx} $ . Let us use this result in equation (1).
So, we get $ \dfrac{d\left( p \right)}{dx}=\dfrac{d\left( {{z}^{3}} \right)}{dx} $ .
 $ \Rightarrow \dfrac{dp}{dx}=\dfrac{d\left( {{z}^{3}} \right)}{dz}\times \dfrac{dz}{dx} $ ---(2).
We know that $ \dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} $ . Let us use this result in equation (2).
 $ \Rightarrow \dfrac{dp}{dx}=3{{z}^{2}}\times \dfrac{dz}{dx} $ .
Now, we have $ \sin 2x=z\Leftrightarrow z=\sin y $ .
 $ \Rightarrow \dfrac{dp}{dx}=3{{\sin }^{2}}\left( 2x \right)\times \dfrac{d\left( \sin y \right)}{dx} $ ---(3).
Applying chain rule in equation (3), we get $ \dfrac{dp}{dx}=3{{\sin }^{2}}\left( 2x \right)\times \dfrac{d\left( \sin y \right)}{dy}\times \dfrac{dy}{dx} $ ---(4).
We know that $ \dfrac{d}{dx}\left( \sin x \right)=\cos x $ . Let us use this result in equation (4).
 $ \Rightarrow \dfrac{dp}{dx}=3{{\sin }^{2}}\left( 2x \right)\times \left( \cos y \right)\times \dfrac{dy}{dx} $ .
Now, we have $ y=2x $ .
 $ \Rightarrow \dfrac{dp}{dx}=3{{\sin }^{2}}\left( 2x \right)\times \left( \cos \left( 2x \right) \right)\times \dfrac{d\left( 2x \right)}{dx} $ ---(5).
We know that $ \dfrac{d\left( ax \right)}{dx}=a $ . Let us use this result in equation (5).
 $ \Rightarrow \dfrac{dp}{dx}=3{{\sin }^{2}}\left( 2x \right)\times \left( \cos \left( 2x \right) \right)\times 2 $ .
 $ \Rightarrow \dfrac{dp}{dx}=6{{\sin }^{2}}\left( 2x \right)\cos \left( 2x \right) $ .
So, we have found the derivative of the function $ {{\sin }^{3}}\left( 2x \right) $ as $ 6{{\sin }^{2}}\left( 2x \right)\cos \left( 2x \right) $ .

Note:
 We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully to avoid confusion and calculation mistakes. Whenever we get this type of problem, we first try to make use of the chain rule of differentiation to get the solution. We can also solve this problem by making use of the fact that $ {{f}^{'}}\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} $ . Similarly, we can expect problems to find the derivative of $ \log \left( \cos \left( 7x \right) \right) $ .