Solveeit Logo
Login

Question

Question: If \(y={{\sin }^{n}}x\cos nx,\) then \(\dfrac{dy}{dx}\) is equal to a). \(n{{\sin }^{n-1}}x\sin \l...

If y=sin⁡nxcos⁡nx,y={{\sin }^{n}}x\cos nx, then dydx\dfrac{dy}{dx} is equal to
a). nsin⁡n−1xsin⁡(n+1)xn{{\sin }^{n-1}}x\sin \left( n+1 \right)x
b). nsin⁡n−1xcos⁡(n−1)xn{{\sin }^{n-1}}x\cos \left( n-1 \right)x
c). nsin⁡n−1xcos⁡nxn{{\sin }^{n-1}}x\cos nx
d). nsin⁡n−1xcos⁡(n+1)xn{{\sin }^{n-1}}x\cos \left( n+1 \right)x

Explanation

Solution

We need to differentiate the trigonometric function present in the form of multiplication of two terms. So we will use the formula in which we can differentiate two terms which are in multiplication. As the formula is d(ab)dx=adbdx+bdadx\dfrac{d\left( ab \right)}{dx}=a\dfrac{db}{dx}+b\dfrac{da}{dx}, in which ‘a’ and ‘b’ are two different terms.

Complete step-by-step solution:
Moving ahead with the question in a stepwise manner, we have y=sin⁡nxcos⁡nx,y={{\sin }^{n}}x\cos nx, and we want to find its first derivative.
So let us use the differentiation formula which allows us to differentiate the function present in the form of multiplication. According to the formula when the function ‘a ×\times b’ is there then we can get its differentiation as first term multiplied by first derivative of second term plus second term multiplied by derivative of first term, as d(ab)dx=adbdx+bdadx\dfrac{d\left( ab \right)}{dx}=a\dfrac{db}{dx}+b\dfrac{da}{dx}.
So by using the same formula let us find out the first derivative of function y=sin⁡nxcos⁡nx,y={{\sin }^{n}}x\cos nx,
So by differentiating according to formula we will get;
y=sin⁡nxcos⁡nx dydx=dsin⁡nxcos⁡nxdx dydx=sin⁡nxdcos⁡nxdx+cos⁡nxdsin⁡nxdx \begin{aligned} & y={{\sin }^{n}}x\cos nx \\\ & \dfrac{dy}{dx}=\dfrac{d{{\sin }^{n}}x\cos nx}{dx} \\\ & \dfrac{dy}{dx}={{\sin }^{n}}x\dfrac{d\cos nx}{dx}+\cos nx\dfrac{d{{\sin }^{n}}x}{dx} \\\ \end{aligned}
Since by chain rule of differentiation we know that differentiation of cos⁡nx\cos nxis−nsin⁡nx-n\sin nx, and differentiation of sin⁡nx{{\sin }^{n}}x is equal to nsin⁡n−1xcos⁡xn{{\sin }^{n-1}}x\cos x. So by putting these value we will get;
dydx=sin⁡nx(−nsin⁡nx)+cos⁡nx(nsin⁡n−1xcos⁡x)\dfrac{dy}{dx}={{\sin }^{n}}x\left( -n\sin nx \right)+\cos nx\left( n{{\sin }^{n-1}}x\cos x \right)
On simplifying it we will get;
dydx=−nsin⁡nxsin⁡nx+ncos⁡nxsin⁡n−1xcos⁡x dydx=nsin⁡n−1x(cos⁡nxcos⁡x−sin⁡xsin⁡nx) \begin{aligned} & \dfrac{dy}{dx}=-n{{\sin }^{n}}x\sin nx+n\cos nx{{\sin }^{n-1}}x\cos x \\\ & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\left( \cos nx\cos x-\sin x\sin nx \right) \\\ \end{aligned}
Since by trigonometric identity we know that cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos (a+b)=\cos a\cos b-\sin a\sin b, so by using the same identity incos⁡nxcos⁡x−sin⁡xsin⁡nx\cos nx\cos x-\sin x\sin nxin above equation we can write it as;
dydx=nsin⁡n−1x(cos⁡nxcos⁡x−sin⁡xsin⁡nx) dydx=nsin⁡n−1x(cos⁡(nx+x)) dydx=nsin⁡n−1xcos⁡(n+1)x \begin{aligned} & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\left( \cos nx\cos x-\sin x\sin nx \right) \\\ & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\left( \cos \left( nx+x \right) \right) \\\ & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\cos (n+1)x \\\ \end{aligned}
So we got nsin⁡n−1xcos⁡(n+1)xn{{\sin }^{n-1}}x\cos (n+1)x.
Hence the correct answer is nsin⁡n−1xcos⁡(n+1)xn{{\sin }^{n-1}}x\cos (n+1)x i.e., option ‘d’ is correct.

Note: First derivative is the differentiation of term or function at once and represented as dydx\dfrac{dy}{dx}. Moreover the identity we used in last cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos (a+b)=\cos a\cos b-\sin a\sin b is the identity of trigonometric function cos⁡\cos which we had used as, we are given with RHS side of identity and we had replaced it with LHS side of the identity.