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Question: Find the derivative of the following function: \({x^4}\left( {5\sin x - 3\cos x} \right)\)...

Find the derivative of the following function:
x4(5sin⁡x−3cos⁡x){x^4}\left( {5\sin x - 3\cos x} \right)

Explanation

Solution

Hint: In this question apply the product rule of differentiation which is given as ddx(uv)=uddxv+vddxu\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u later on in the solution apply the differentiation property of sin x, cos x and xn{x^n} which is given as ddx(sin⁡x)=cos⁡x and ddx(cos⁡x)=−sin⁡x and ddxxn=nxn−1\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x{\text{ and }}\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let
y=x4(5sin⁡x−3cos⁡x)y = {x^4}\left( {5\sin x - 3\cos x} \right)
Now differentiate it w.r.t. x we have,
⇒ddxy=ddx[x4(5sin⁡x−3cos⁡x)]\Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{x^4}\left( {5\sin x - 3\cos x} \right)} \right]
Now here we use product rule of differentiate which is given as
ddx(uv)=uddxv+vddxu\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u so use this property in above equation we have,
⇒ddxy=x4ddx(5sin⁡x−3cos⁡x)+(5sin⁡x−3cos⁡x)ddxx4\Rightarrow \dfrac{d}{{dx}}y = {x^4}\dfrac{d}{{dx}}\left( {5\sin x - 3\cos x} \right) + \left( {5\sin x - 3\cos x} \right)\dfrac{d}{{dx}}{x^4}
Now as we know that differentiation of ddx(sin⁡x)=cos⁡x and ddx(cos⁡x)=−sin⁡x and ddxxn=nxn−1\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x{\text{ and }}\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} so use this property in above equation we have,
⇒ddxy=x4(5cos⁡x+3sin⁡x)+(5sin⁡x−3cos⁡x)4x3\Rightarrow \dfrac{d}{{dx}}y = {x^4}\left( {5\cos x + 3\sin x} \right) + \left( {5\sin x - 3\cos x} \right)4{x^3}
Now simplify this equation we have,
⇒ddxy=5x4cos⁡x+3x4sin⁡x+20x3sin⁡x−12x3cos⁡x\Rightarrow \dfrac{d}{{dx}}y = 5{x^4}\cos x + 3{x^4}\sin x + 20{x^3}\sin x - 12{x^3}\cos x
⇒ddxy=x3cos⁡x(5x−12)+x3sin⁡x(3x+20)\Rightarrow \dfrac{d}{{dx}}y = {x^3}\cos x\left( {5x - 12} \right) + {x^3}\sin x\left( {3x + 20} \right)
So this is the required differentiation.

Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of sin x, cos x and xn{x^n} differentiation which is stated above then first apply the product rule as above then use the property of differentiation of sin x, cos x and xn{x^n} as above and simplify we will get the required answer.