If f(x) = |cos x - sin x|,then f^{prime}left( | vedclass

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

Question

(A) Given the function $f(x) = |\cos x - \sin x|$.In the neighborhood of $x = \frac{\pi}{6}$,we have $\cos x > \sin x$,so $f(x) = \cos x - \sin x$.Now

Vedclass · Accepted Answer

$-\frac{1}{2}(1+\sqrt{3})$

Vedclass · Answer

(A) Given the function $f(x) = |\cos x - \sin x|$.In the neighborhood of $x = \frac{\pi}{6}$,we have $\cos x > \sin x$,so $f(x) = \cos x - \sin x$.Now,differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x$.Substitute $x = \frac{\pi}{6}$ into the derivative:$f^{\prime}\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right)$.Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,we get:$f^{\prime}\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(1 + \sqrt{3})$.

If $f(x) = |\cos x - \sin x|$,then $f^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

Similar Questions

At $x = \sqrt{\frac{\pi}{2}}$,find the value of $\frac{d}{dx} \cos(\sin(x^2))$.

Suppose that $h(x) = f(x) \cdot g(x)$ and $F(x) = f(g(x))$,where $f(2) = 3$,$g(2) = 5$,$g'(2) = 4$,$f'(2) = -2$,and $f'(5) = 11$. Then:

The differential coefficient of $f(\sin x)$ with respect to $x,$ where $f(x) = \log x,$ is

Find the derivative of the function: $f(x) = 5 \sin x - 6 \cos x + 7$.

Evaluate: $\frac{d}{d x}\left[e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module