Question

If $y={{\sin }^{n}}x\cos nx,$ then $\dfrac{dy}{dx}$ is equal to
a). $n{{\sin }^{n-1}}x\sin \left( n+1 \right)x$
b). $n{{\sin }^{n-1}}x\cos \left( n-1 \right)x$
c). $n{{\sin }^{n-1}}x\cos nx$
d). $n{{\sin }^{n-1}}x\cos \left( n+1 \right)x$

Answer 1

Hint: 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 $\dfrac{d\left( ab \right)}{dx}=a\dfrac{db}{dx}+b\dfrac{da}{dx}$, 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 $y={{\sin }^{n}}x\cos nx,$ 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 $\dfrac{d\left( ab \right)}{dx}=a\dfrac{db}{dx}+b\dfrac{da}{dx}$.
So by using the same formula let us find out the first derivative of function $y={{\sin }^{n}}x\cos nx,$
So by differentiating according to formula we will get;
$\begin{align}
  & y={{\sin }^{n}}x\cos nx \\
 & \dfrac{dy}{dx}=\dfrac{d{{\sin }^{n}}x\cos nx}{dx} \\
 & \dfrac{dy}{dx}={{\sin }^{n}}x\dfrac{d\cos nx}{dx}+\cos nx\dfrac{d{{\sin }^{n}}x}{dx} \\
\end{align}$
Since by chain rule of differentiation we know that differentiation of $\cos nx$is$-n\sin nx$, and differentiation of ${{\sin }^{n}}x$ is equal to $n{{\sin }^{n-1}}x\cos x$. So by putting these value we will get;
$\dfrac{dy}{dx}={{\sin }^{n}}x\left( -n\sin nx \right)+\cos nx\left( n{{\sin }^{n-1}}x\cos x \right)$
On simplifying it we will get;
$\begin{align}
  & \dfrac{dy}{dx}=-n{{\sin }^{n}}x\sin nx+n\cos nx{{\sin }^{n-1}}x\cos x \\
 & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\left( \cos nx\cos x-\sin x\sin nx \right) \\
\end{align}$
Since by trigonometric identity we know that $\cos (a+b)=\cos a\cos b-\sin a\sin b$, so by using the same identity in$\cos nx\cos x-\sin x\sin nx$in above equation we can write it as;
$\begin{align}
  & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\left( \cos nx\cos x-\sin x\sin nx \right) \\
 & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\left( \cos \left( nx+x \right) \right) \\
 & \dfrac{dy}{dx}=n{{\sin }^{n-1}}x\cos (n+1)x \\
\end{align}$
So we got $n{{\sin }^{n-1}}x\cos (n+1)x$.
Hence the correct answer is $n{{\sin }^{n-1}}x\cos (n+1)x$ i.e., option ‘d’ is correct.

Note: First derivative is the differentiation of term or function at once and represented as $\dfrac{dy}{dx}$. Moreover the identity we used in last $\cos (a+b)=\cos a\cos b-\sin a\sin b$ is the identity of trigonometric function $\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.