Solveeit Logo

Question

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

How do you find the derivative of sin7(x){{\sin }^{7}}\left( x \right)?

Explanation

Solution

Assume the function sinx\sin x as f(x)f\left( x \right), so that the given function gets converted into the form (f(x))n{{\left( f\left( x \right) \right)}^{n}} where n = 7. Now, differentiate the function with respect to the value x and use the formula d[f(x)n]dx=n×(f(x))n1×f(x)\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 d(sinx)dx=cosx\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 sin7(x){{\sin }^{7}}\left( x \right) and we are asked to differentiate it. Let us assume the given function the as y. So, we have,
f(x)=sinx\Rightarrow f\left( x \right)=\sin x
Therefore, assuming the given function as y, we have,
y=(f(x))7\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=(f(x))ny={{\left( f\left( x \right) \right)}^{n}}, where n = 7, whose derivative is given by the power reduction formula given as: d[f(x)n]dx=n×(f(x))n1×f(x)\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,
dydx=7×(sinx)71×d[sinx]dx dydx=7×(sinx)6×d[sinx]dx dydx=7×sin6x×d[sinx]dx \begin{aligned} & \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{aligned}
Using the basic formula of the derivative of the sine function given as: d(sinx)dx=cosx\dfrac{d\left( \sin x \right)}{dx}=\cos x, we get,
dydx=7×sin6x×cosx dydx=7sin6xcosx \begin{aligned} & \Rightarrow \dfrac{dy}{dx}=7\times {{\sin }^{6}}x\times \cos x \\\ & \Rightarrow \dfrac{dy}{dx}=7{{\sin }^{6}}x\cos x \\\ \end{aligned}
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, uv\dfrac{u}{v} rule etc. as they make our question easy to solve. Remember the derivatives of some common functions like: - xn,ex{{x}^{n}},{{e}^{x}}, trigonometric functions, inverse trigonometric functions, logarithmic functions etc. as they are used frequently in calculus.