Question

If $y={\cos }^{n}x.\sin nx$and $\frac{dy}{dx}=n.{\cos }^{n-1}x\times \cos B$ then $B$ is equals to ?

• A. $(n–1)x$
• B. $(n+1)x$
• C. $nx$
• D. $(1–n)x$

Answer 1

The correct option is B

$(n+1)x$

Explanation for the correct option:

Find the value of $B$:

Given functions,

$y={\cos }^{n}x.\sin nx$

Differentiate the function $y={\cos }^{n}x·\sin nx$ with respect to $x$:

$\begin{array}{rcl}\frac{dy}{dx}& =& n·co{s}^{n–1}x·(–sinx)·sinnx+ncosnx·co{s}^{n}x\left(n\right)\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{,}\mathbf{}\frac{\mathbf{d}\mathbf{\left(}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{]}\\ & =& nco{s}^{n–1}x·sinnx·(–sinx)+nco{s}^{n}xcosnx\\ & =& nco{s}^{n}x\left[\frac{(–\sin x·\sin nx)}{\left(\cos x\right)}+\cos nx\right]\\ & =& nco{s}^{n}x\left[\frac{(–\sin x\sin nx+\cos x\cos nx)}{\cos x}\right]\\ & =& n·co{s}^{n–1}x·cos(x+nx)\end{array}$

Compare the obtained result with $\frac{dy}{dx}=n.{\cos }^{n-1}x\times \cos B$

$\begin{array}{rcl}B& =& x+nx\\ & =& x(n+1)\end{array}$

Hence, the option (B) is the correct.