Solveeit Logo
Login

Question

Question: What is the derivative of \({{\sin }^{5}}\left( x \right)\)?...

What is the derivative of sin⁡5(x){{\sin }^{5}}\left( x \right)?

Explanation

Solution

For solving this question you should know about the differentiation of trigonometric functions and we can solve this by this. Here we can see the two terms that we are seeing, so we will differentiate the outside term first and then we differentiate the inside term. Thus, we will get the answer for this.

Complete step by step solution:
According to our question we have to calculate the derivative of sin⁡5(x){{\sin }^{5}}\left( x \right). As we know that the differentiation of trigonometric functions will always give us trigonometric values and if we differentiate to inverse trigonometric functions then it will give any other value than trigonometric functions. And by the differentiation of any variable, it will reduce the power of the variable by 1 and the power will multiply by it again. So, according to our question,
y=sin⁡5(x)y={{\sin }^{5}}\left( x \right)
If we consider,
y=(f(x))5=(sin⁡(x))5y={{\left( f\left( x \right) \right)}^{5}}={{\left( \sin \left( x \right) \right)}^{5}}
Then if we differentiate it, we will get as,
y=(f(x))n⇒dydx=n.f′(x).f(x)n−1y={{\left( f\left( x \right) \right)}^{n}}\Rightarrow \dfrac{dy}{dx}=n.f'\left( x \right).f{{\left( x \right)}^{n-1}}
So, according to the given question, we can write it as,
y=(sin⁡(x))5y={{\left( \sin \left( x \right) \right)}^{5}}
We will differentiate this using the chain rule. So, we have,
[dydx=dydu×dudx]………(A)\left[ \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \right]\ldots \ldots \ldots \left( A \right)
Let u=sin⁡x⇒dudx=cos⁡xu=\sin x\Rightarrow \dfrac{du}{dx}=\cos x
And let y=u5⇒dydu=5u4y={{u}^{5}}\Rightarrow \dfrac{dy}{du}=5{{u}^{4}}
We will now substitute these values in expression A, and convert uu back to xx, so,
⇒dydx=5u4(cos⁡x)=5sin⁡4xcos⁡x\Rightarrow \dfrac{dy}{dx}=5{{u}^{4}}\left( \cos x \right)=5{{\sin }^{4}}x\cos x
So, the derivative of sin⁡5(x){{\sin }^{5}}\left( x \right) is 5sin⁡4xcos⁡x5{{\sin }^{4}}x\cos x.

Note: For calculating the differentiation of any trigonometric function we have to learn all the trigonometric formulas for differentiation because this is completely based on formulas. And always we should be careful about the power of variables and specially if they are the variables whose regarding we are differentiating that.