Question

If $y = {\cos ^n}x.\sin nx$and $\left( {\dfrac{{dy}}{{dx}}} \right) = n.{\cos ^{n - 1}}x.\cos B$then $ B = $
1. $\left( {n - 1} \right)x$
2. $\left( {n + 1} \right)x$
3. $nx$
4. $\left( {1 - n} \right)x$

Answer 1

Hint: In this question, we are given that $y = {\cos ^n}x.\sin nx$ and we have to find the value of $B$. The first step is to differentiate the given function with respect to $x$. Then, apply the product rule as the function is in product form. After that, you’ll get the function inside the function for that apply the chain rule, and solve further. Convert the final answer similar to $\left( {\dfrac{{dy}}{{dx}}} \right) = n.{\cos ^{n - 1}}x.\cos B$ this given form. In the last compare them and you will get the value of $B$.

Formula Used:
Product rule –
$\dfrac{d}{{dx}}\left( {pq} \right) = p\dfrac{{dq}}{{dx}} + q\dfrac{{dp}}{{dx}}$
Chain rule –
$\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \times \dfrac{d}{{dx}}g\left( x \right)$
Trigonometric identity –
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$

Complete step by step Solution:
Given that,
$y = {\cos ^n}x.\sin nx - - - - - \left( 1 \right)$
$\left( {\dfrac{{dy}}{{dx}}} \right) = n.{\cos ^{n - 1}}x.\cos B - - - - - \left( 2 \right)$
Differentiate equation (1) with respect to $x$,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\cos }^n}x.\sin nx} \right)$
Applying product rule of differentiation, $\dfrac{d}{{dx}}\left( {pq} \right) = p\dfrac{{dq}}{{dx}} + q\dfrac{{dp}}{{dx}}$
$\dfrac{{dy}}{{dx}} = {\cos ^n}x\dfrac{d}{{dx}}\left( {\sin nx} \right) + \sin nx\dfrac{d}{{dx}}\left( {{{\cos }^n}x} \right)$
Now, applying chain rule of differentiation $\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \times \dfrac{d}{{dx}}g\left( x \right)$
$\dfrac{{dy}}{{dx}} = {\cos ^n}x\left( {\cos nx} \right)\left( n \right) + \sin nx\left( {n{{\cos }^{n - 1}}x} \right)\left( { - \sin x} \right)$
$\dfrac{{dy}}{{dx}} = n{\cos ^n}x\left( {\cos nx - \dfrac{{\sin x\sin nx}}{{\cos x}}} \right)$
$\dfrac{{dy}}{{dx}} = n{\cos ^n}x\left( {\dfrac{{\cos x\cos nx - \sin x\sin nx}}{{\cos x}}} \right)$
$\dfrac{{dy}}{{dx}} = n{\cos ^{n - 1}}x\left( {\cos x\cos nx - \sin x\sin nx} \right)$
Using trigonometric identity, $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
$\dfrac{{dy}}{{dx}} = n.{\cos ^{n - 1}}x.\cos \left( {x + nx} \right) - - - - - \left( 3 \right)$
Comparing equations (2) and (3), we get
$B = x + nx$
$B = x\left( {n + 1} \right)$

Hence, the correct option is 2.

Note: The key concept involved in solving this problem is a good knowledge of trigonometry rules, and identities. Students must remember that to solve such problems you should know when to apply which formula and while solving any function you are getting any terms whether they are similar to any identity or not. If yes, convert them into the simplest form. For example, in this question product rule and chain rule are applied. If you’ll be unaware of the chain rule and differentiate directly answer will be wrong.