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Question: Find derivative of \[y = {\sin ^n}x\cos nx\]. A. \[n{\sin ^{n - 1}}x\cos (n + 1)x\] B. \[n{\sin ...

Find derivative of y=sin⁡nxcos⁡nxy = {\sin ^n}x\cos nx.
A. nsin⁡n−1xcos⁡(n+1)xn{\sin ^{n - 1}}x\cos (n + 1)x
B. nsin⁡n−1xcos⁡nxn{\sin ^{n - 1}}x\cos nx
C. nsin⁡n−1xcos⁡(n−1)xn{\sin ^{n - 1}}x\cos (n - 1)x
D. nsin⁡n−1xsin⁡(n+1)xn{\sin ^{n - 1}}x\sin (n + 1)x

Explanation

Solution

We can find the solution for the given problem by using the concept of differentiation. and Moreover, the given problem is of the form of product of two functions so we have to use the product rule of differentiation. and rearranging the terms we will get the required solution.

Complete step by step answer:
Now let us consider the given function which we have to differentiate as y
That is y=sin⁡nxcos⁡nxy = {\sin ^n}x\cos nx
Since y is of the form of product of two functions so apply the product rule and differentiate by taking sin⁡nx{\sin ^n}xas first function and cos⁡nx\cos nxas second function, we get

dydx = - sinnx sinnx(n) + cosnx nsinn - 1x cosx\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = - si}}{{\text{n}}^{\text{n}}}{\text{x sinnx(n) + cosnx nsi}}{{\text{n}}^{{\text{n - 1}}}}{\text{x cosx}}
(By product and chain rule)
dydx = cosnx nsinn - 1x cosx - sinnx sinnx(n)\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cosnx nsi}}{{\text{n}}^{{\text{n - 1}}}}{\text{x cosx - si}}{{\text{n}}^{\text{n}}}{\text{x sinnx(n)}}
Now let us multiply and divide the second term by sinx{\text{ sinx}}we get
dydx = cosnx nsinn - 1x cosx - sinnx sin⁡xsin⁡xsinnx(n)\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cosnx nsi}}{{\text{n}}^{{\text{n - 1}}}}{\text{x cosx - si}}{{\text{n}}^{\text{n}}}{\text{x }}\dfrac{{\sin x}}{{\sin x}}{\text{sinnx(n)}}
Now we can rewrite the second term as
dydx = cosnx n sinn - 1x cosx - sinn - 1x sinx sinnx(n)\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = cosnx n si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x cosx - si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x sinx sinnx(n)}}
Now taking n sinn - 1x{\text{n si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}as common term we get
dydx = n sinn - 1x(cosnx cosx - sinx sinnx)\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = n si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}({\text{cosnx cosx - sinx sinnx)}}
The term in the bracket is of the form cos⁡(a+b)\cos (a + b)so rewriting the terms in the bracket we get
dydx = n sinn - 1x(cos(nx + x))\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = n si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}({\text{cos(nx + x))}}
Taking x as a common term we get

dydx = n sinn - 1x(cos(n + 1)x)\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = n si}}{{\text{n}}^{{\text{n - 1}}}}{\text{x}}({\text{cos(n + 1)x)}}
So, the correct answer is “Option A”.

Note: In calculus, the product rule is a formula used to find the derivatives of products of two or more functions.
The chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g.