Question

What is the derivative of ${{\sin }^{4}}x$?

Answer 1

Hint: First of all write the given function as ${{\sin }^{4}}x={{\left( \sin x \right)}^{4}}$. Assume the function $\left( \sin x \right)$ as $f\left( x \right)$ and write \[{{\left( \sin x \right)}^{4}}\] as ${{\left[ f\left( x \right) \right]}^{4}}$ . Now, use the chain rule of differentiation to differentiate the function. First differentiate the function ${{\left[ f\left( x \right) \right]}^{4}}$ with respect to the function $f\left( x \right)$ and then differentiate the function $f\left( x \right)$ with respect to x. Finally, take the product of these two derivatives to get the answer. Use the formulas $\dfrac{d\left[ {{\left( f\left( x \right) \right)}^{n}} \right]}{d\left[ f\left( x \right) \right]}=n{{\left( f\left( x \right) \right)}^{n-1}}$ and $\dfrac{d\left[ \sin x \right]}{dx}=\cos x$ to get the answer.

Complete step by step solution:
Here we have been provided with the function ${{\sin }^{4}}x$ and we are asked to find its derivative. Here we will use the chain rule of derivatives to get the answer.
Now, we can write the given function as ${{\sin }^{4}}x={{\left( \sin x \right)}^{4}}$. Assuming the function $\left( \sin x \right)$ as $f\left( x \right)$ we have the function \[{{\left( \sin x \right)}^{4}}\] of the form ${{\left[ f\left( x \right) \right]}^{4}}$. So we have,
$\Rightarrow {{\left( \sin x \right)}^{4}}={{\left[ f\left( x \right) \right]}^{4}}$
On differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=\dfrac{d{{\left[ f\left( x \right) \right]}^{4}}}{dx}$
Now, according to the chain rule of differentiation first we have to differentiate the function ${{\left[ f\left( x \right) \right]}^{4}}$ with respect to $f\left( x \right)$ and then we have to differentiate $f\left( x \right)$ with respect to x. Finally, we need to consider their product to get the relation. So we get,
$\Rightarrow \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=\dfrac{d{{\left[ f\left( x \right) \right]}^{4}}}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$
Using the formula $\dfrac{d\left[ {{\left( f\left( x \right) \right)}^{n}} \right]}{d\left[ f\left( x \right) \right]}=n{{\left( f\left( x \right) \right)}^{n-1}}$ we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=4{{\left[ f\left( x \right) \right]}^{4-1}}\times \dfrac{d\left[ f\left( x \right) \right]}{dx} \\
 & \Rightarrow \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=4{{\left[ f\left( x \right) \right]}^{3}}\times \dfrac{d\left[ f\left( x \right) \right]}{dx} \\
\end{align}\]
Substituting the value of $f\left( x \right)$ we get,
$\Rightarrow \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=4{{\left[ \sin x \right]}^{3}}\times \dfrac{d\left[ \sin x \right]}{dx}$
Using the formula $\dfrac{d\left[ \sin x \right]}{dx}=\cos x$ we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=4{{\left[ \sin x \right]}^{3}}\times \cos x \\
 & \therefore \dfrac{d\left[ {{\left( \sin x \right)}^{4}} \right]}{dx}=4{{\sin }^{3}}x\cos x \\
\end{align}\]
Hence, the above relation is our answer.

Note: You must remember all the basic rules and formulas of differentiation like: - product rule, chain rule, \[\dfrac{u}{v}\] rule etc. Remember the formulas of derivatives of all the basic functions like trigonometric and inverse trigonometric functions, logarithmic functions, exponential functions etc. In case if the argument of the trigonometric functions is linear like $\sin \left( ax+b \right)$ then the formula becomes $\dfrac{d\left( \sin \left( ax+b \right) \right)}{dx}=a\cos \left( ax+b \right)$. This is due to the chain rule of derivatives.