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

If $f(x)$ is a polynomial such that $f(x) = [f'(x)]^2$ and $f(2) = 0$, then $f(-2) = \dots$

Find the value of $k$ if $\frac{d}{d x}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos (4 x)}}}}\right\} = k \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

Differentiate $a^{x}$ with respect to $x$,where $a$ is a positive constant.

$x = \frac{1-\sqrt{y}}{1+\sqrt{y}} \Rightarrow \frac{dy}{dx}$ is equal to

If $y = \frac{5x}{\sqrt[3]{(1 - x)^2}} + \cos^2(2x + 1)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module