If f(x)=frac{cos ^2 x}{1+sin ^2 x}, then flef | vedclass

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

Question

(D) Given $f(x)=\frac{\cos ^2 x}{1+\sin ^2 x}$.First, evaluate $f\left(\frac{\pi}{4}\right)$:$f\left(\frac{\pi}{4}\right)=\frac{\cos ^2(\pi/4)}{1+\sin

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given $f(x)=\frac{\cos ^2 x}{1+\sin ^2 x}$.First, evaluate $f\left(\frac{\pi}{4}ight)$:$f\left(\frac{\pi}{4}ight)=\frac{\cos ^2(\pi/4)}{1+\sin ^2(\pi/4)}=\frac{(1/\sqrt{2})^2}{1+(1/\sqrt{2})^2}=\frac{1/2}{1+1/2}=\frac{1/2}{3/2}=\frac{1}{3}$.Now, differentiate $f(x)$ using logarithmic differentiation:$\ln(f(x)) = 2\ln(\cos x) - \ln(1+\sin^2 x)$.Differentiating with respect to $x$:$\frac{f'(x)}{f(x)} = 2\left(\frac{-\sin x}{\cos x}ight) - \frac{2\sin x \cos x}{1+\sin^2 x} = -2	an x - \frac{\sin 2x}{1+\sin^2 x}$.At $x = \frac{\pi}{4}$:$\frac{f'(\pi/4)}{f(\pi/4)} = -2	an(\pi/4) - \frac{\sin(\pi/2)}{1+\sin^2(\pi/4)} = -2(1) - \frac{1}{1+1/2} = -2 - \frac{1}{3/2} = -2 - \frac{2}{3} = -\frac{8}{3}$.Since $f(\pi/4) = 1/3$, we have $f'(\pi/4) = -\frac{8}{3} 	imes \frac{1}{3} = -\frac{8}{9}$.Finally, calculate $f(\pi/4) - 3f'(\pi/4)$:$\frac{1}{3} - 3\left(-\frac{8}{9}ight) = \frac{1}{3} + \frac{8}{3} = \frac{9}{3} = 3$.

If $f(x)=\frac{\cos ^2 x}{1+\sin ^2 x}$, then $f\left(\frac{\pi}{4}\right)-3 f^{\prime}\left(\frac{\pi}{4}\right)=$

Similar Questions

If $y = x^x + x^{\frac{1}{x}}$,then $\frac{dy}{dx}$ is equal to

If $x^p \cdot y^q = (x + y)^{p + q}$,then $\frac{dy}{dx}$ is:

If $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$, then $\frac{d y}{d x}=$

Differentiate the function with respect to $x$: $x^{x \cos x} + \frac{x^{2}+1}{x^{2}-1}$

If $y = x^{\sqrt{x}}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module