The first derivative of the function left[ co | vedclass

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

Question

(A) Let $f(x) = \cos^{-1}\left( \sin \sqrt{\frac{1+x}{2}} ight) + x^x$. Using the identity $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we have: $f(

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(A) Let $f(x) = \cos^{-1}\left( \sin \sqrt{\frac{1+x}{2}} ight) + x^x$. Using the identity $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we have: $f(x) = \cos^{-1}\left[ \cos \left( \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} ight) ight] + x^x$. Simplifying the expression: $f(x) = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$. Differentiating with respect to $x$: $f'(x) = 0 - \frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + \frac{d}{dx}(x^x)$. Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we get: $f'(x) = -\frac{1}{2\sqrt{2(1+x)}} + x^x(1 + \ln x)$. At $x = 1$: $f'(1) = -\frac{1}{2\sqrt{2(1+1)}} + 1^1(1 + \ln 1) = -\frac{1}{2\sqrt{4}} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

The first derivative of the function $\left[ \cos^{-1}\left( \sin \sqrt{\frac{1+x}{2}} \right) + x^x \right]$ with respect to $x$ at $x = 1$ is

Similar Questions

Find the derivative of $f(x) = 1 + x + x^{2} + x^{3} + \dots + x^{50}$ at $x = 1$.

If $y = \sin^{-1}\sqrt{1 - x} + \cos^{-1}\sqrt{x}$,then $\frac{dy}{dx} = $

$\frac{d}{dx}\sqrt{\sec^2 x + \text{cosec}^2 x} = $

The differential coefficient of the function $|x - 1| + |x - 3|$ at the point $x = 2$ is

$\frac{d}{dx} \left( \sqrt{3} \sin \left(2x + \frac{\pi}{3} \right) + \cos \left(2x + \frac{\pi}{3} \right) \right) = $ . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module