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

For the function $f(x) = x^2 - 6x + 8$ where $2 \le x \le 4$,the value of $x$ for which $f'(x)$ vanishes is:

Find the derivative: $\frac{d}{dx} \left( \frac{1}{x^4 \sec x} \right) = $

If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$,then $f'(a) = $

Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{a+b \sin x}{c+d \cos x}$

Which one of the following is false?

Vedclass Test Series

Exam Paper Generator

Online Exam Module