If y = sin^{98}(x) cdot cos^{39}(x),then find | vedclass

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

Question

(C) Given $y = \sin^{98} x \cdot \cos^{39} x$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin

Vedclass · Accepted Answer

$\left(98 \cos^{99} x \cdot \sin^{38} x\right) - \left(39 \sin^{40} x \cdot \cos^{97} x\right)$

Vedclass · Answer

(C) Given $y = \sin^{98} x \cdot \cos^{39} x$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin^{98} x \cdot \frac{d}{dx}(\cos^{39} x) + \cos^{39} x \cdot \frac{d}{dx}(\sin^{98} x)$ $\frac{dy}{dx} = \sin^{98} x \cdot (39 \cos^{38} x \cdot (-\sin x)) + \cos^{39} x \cdot (98 \sin^{97} x \cdot \cos x)$ $\frac{dy}{dx} = -39 \sin^{99} x \cdot \cos^{38} x + 98 \sin^{97} x \cdot \cos^{40} x$ Rearranging the terms: $\frac{dy}{dx} = 98 \sin^{97} x \cdot \cos^{40} x - 39 \sin^{99} x \cdot \cos^{38} x$. Note: The provided options seem to have typographical errors in the exponents. Based on the derivation,the correct expression is $98 \sin^{97} x \cos^{40} x - 39 \sin^{99} x \cos^{38} x$.

If $y = \sin^{98}(x) \cdot \cos^{39}(x)$,then find $\frac{dy}{dx}$.

Similar Questions

$\frac{d}{dx} \log_{\sqrt{x}} \left(\frac{1}{x}\right)$ is equal to

The derivative of $y = (1-x)(2-x)(3-x) \dots (n-x)$ at $x=1$ is

Find the derivative of $\frac{x^{n}-a^{n}}{x-a}$ with respect to $x$,where $a$ is a constant.

If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$,then $\frac{dy}{dx} = $

If $f(x) = x \tan^{-1} x$,then $f'(1) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module