Question

How do you find the derivative of \[{{\sin }^{7}}\left( x \right)\]?

Answer 1

Hint: Assume the function $\sin x$ as \[f\left( x \right)\], so that the given function gets converted into the form ${{\left( f\left( x \right) \right)}^{n}}$ where n = 7. Now, differentiate the function with respect to the value x and use the formula $\dfrac{d\left[ f{{\left( x \right)}^{n}} \right]}{dx}=n\times {{\left( f\left( x \right) \right)}^{n-1}}\times f'\left( x \right)$ where f’(x) is the derivative of the assumed function f(x). Use the basic formula $\dfrac{d\left( \sin x \right)}{dx}=\cos x$ to get the answer.

Complete step by step solution:
Here, we have been provided with the function \[{{\sin }^{7}}\left( x \right)\] and we are asked to differentiate it. Let us assume the given function the as y. So, we have,
\[\Rightarrow f\left( x \right)=\sin x\]
Therefore, assuming the given function as y, we have,
\[\Rightarrow y={{\left( f\left( x \right) \right)}^{7}}\]
So, we have to differentiate the above function. Clearly, we can see that the above function is of the form $y={{\left( f\left( x \right) \right)}^{n}}$, where n = 7, whose derivative is given by the power reduction formula given as: $\dfrac{d\left[ f{{\left( x \right)}^{n}} \right]}{dx}=n\times {{\left( f\left( x \right) \right)}^{n-1}}\times f'\left( x \right)$, where f’(x) is the derivative of f(x),so using this formula we get on differentiating both the sides with respect to the variable x,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=7\times {{\left( \sin x \right)}^{7-1}}\times \dfrac{d\left[ \sin x \right]}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=7\times {{\left( \sin x \right)}^{6}}\times \dfrac{d\left[ \sin x \right]}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=7\times {{\sin }^{6}}x\times \dfrac{d\left[ \sin x \right]}{dx} \\
\end{align}$
Using the basic formula of the derivative of the sine function given as: $\dfrac{d\left( \sin x \right)}{dx}=\cos x$, we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=7\times {{\sin }^{6}}x\times \cos x \\
 & \Rightarrow \dfrac{dy}{dx}=7{{\sin }^{6}}x\cos x \\
\end{align}$
Hence, the above relation is our answer.

Note: You must remember all the basic rules and formulas of differentiation like: - power reduction rule, product rule, chain rule, \[\dfrac{u}{v}\] rule etc. as they make our question easy to solve. Remember the derivatives of some common functions like: - \[{{x}^{n}},{{e}^{x}}\], trigonometric functions, inverse trigonometric functions, logarithmic functions etc. as they are used frequently in calculus.