If f(x)=5 cos ^3 x-3 sin ^2 x and g(x)=4 sin | vedclass

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

Question

(D) Given $f(x) = 5 \cos^3 x - 3 \sin^2 x$ and $g(x) = 4 \sin^3 x + \cos^2 x$. First,find the derivative of $f(x)$ with respect to $x$: $\frac{df}{dx}

Vedclass · Accepted Answer

$-\left(\frac{15 \cos x+6}{12 \sin x-2}\right)$

Vedclass · Answer

(D) Given $f(x) = 5 \cos^3 x - 3 \sin^2 x$ and $g(x) = 4 \sin^3 x + \cos^2 x$. First,find the derivative of $f(x)$ with respect to $x$: $\frac{df}{dx} = 5(3 \cos^2 x)(-\sin x) - 3(2 \sin x \cos x) = -15 \cos^2 x \sin x - 6 \sin x \cos x$. Next,find the derivative of $g(x)$ with respect to $x$: $\frac{dg}{dx} = 4(3 \sin^2 x)(\cos x) + 2 \cos x(-\sin x) = 12 \sin^2 x \cos x - 2 \sin x \cos x$. Now,find the derivative of $f(x)$ with respect to $g(x)$ using the chain rule: $\frac{df}{dg} = \frac{df/dx}{dg/dx} = \frac{-15 \cos^2 x \sin x - 6 \sin x \cos x}{12 \sin^2 x \cos x - 2 \sin x \cos x}$. Factor out $-\sin x \cos x$ from the numerator and $\sin x \cos x$ from the denominator: $\frac{df}{dg} = \frac{-\sin x \cos x (15 \cos x + 6)}{\sin x \cos x (12 \sin x - 2)} = -\frac{15 \cos x + 6}{12 \sin x - 2}$. Thus,the correct option is $D$.

If $f(x)=5 \cos ^3 x-3 \sin ^2 x$ and $g(x)=4 \sin ^3 x+\cos ^2 x$,then the derivative of $f(x)$ with respect to $g(x)$ is

Similar Questions

If $x = a \cos \theta$ and $y = b \sin \theta$,then find the value of $\left[\frac{d^2 y}{d x^2}\right]_{\theta = \frac{\pi}{4}}$.

If $x=3 \sqrt{2} \cos ^3 \theta$ and $y=4 \tan ^2 \theta$, then $\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}} = $

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \cos \theta$ and $y = b \cos \theta$.

If $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$,then $\frac{dy}{dx}=$

If the equation of the normal to the curve $x=\sqrt{t}$,$y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module