frac{d}{dx}[sin^n x cos nx] = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $y = \sin^n x \cos nx$. Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin^n x \frac{d}{dx}(

Vedclass · Accepted Answer

$n \sin^{n-1} x \cos(n+1)x$

Vedclass · Answer

(A) Let $y = \sin^n x \cos nx$. Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin^n x \frac{d}{dx}(\cos nx) + \cos nx \frac{d}{dx}(\sin^n x)$ $\frac{dy}{dx} = \sin^n x (-n \sin nx) + \cos nx (n \sin^{n-1} x \cos x)$ $\frac{dy}{dx} = n \sin^{n-1} x [\cos x \cos nx - \sin x \sin nx]$ Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $\frac{dy}{dx} = n \sin^{n-1} x \cos(nx + x) = n \sin^{n-1} x \cos(n+1)x$.

$\frac{d}{dx}[\sin^n x \cos nx] = $

Similar Questions

The value of $\left(\frac{\Delta^{2}}{E}\right) x^{3}$ at $h=1$ is

Find the derivative of the function given by $f(x)=\sin(x^{2}).$

If $f(x) = x + 2,$ then $f'(f(x))$ at $x = 4$ is

$\frac{d}{d x}\left[\cos ^2\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

Find the derivative of the function $f(x) = x^{4}(5 \sin x - 3 \cos x)$ with respect to $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module