If f(x) = frac{sin(frac{pi x}{4})}{x + 1},the | vedclass

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

Question

(D) Let $L = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$.Using the definition of the derivative,$\lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = f'(1)$.

Vedclass · Accepted Answer

$\frac{\pi - 4}{16\sqrt{2}}$

Vedclass · Answer

(D) Let $L = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$.Using the definition of the derivative,$\lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = f'(1)$.We can rewrite the limit as $\lim_{h 	o 0} \left( \frac{f(1+h) - f(1)}{h} \cdot \frac{1}{h+2} ight) = f'(1) \cdot \frac{1}{2} = \frac{f'(1)}{2}$.Given $f(x) = \frac{\sin(\frac{\pi x}{4})}{x + 1}$,we use the quotient rule: $f'(x) = \frac{(x+1) \cdot \cos(\frac{\pi x}{4}) \cdot \frac{\pi}{4} - \sin(\frac{\pi x}{4}) \cdot 1}{(x+1)^2}$.At $x = 1$,$f'(1) = \frac{(1+1) \cdot \cos(\frac{\pi}{4}) \cdot \frac{\pi}{4} - \sin(\frac{\pi}{4})}{(1+1)^2} = \frac{2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} - \frac{1}{\sqrt{2}}}{4} = \frac{\frac{\pi}{2\sqrt{2}} - \frac{1}{\sqrt{2}}}{4} = \frac{\pi - 2}{8\sqrt{2}}$.Thus,the limit is $\frac{f'(1)}{2} = \frac{\pi - 2}{16\sqrt{2}}$.

If $f(x) = \frac{\sin(\frac{\pi x}{4})}{x + 1}$,then $\lim_{h \to 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$ is -

Similar Questions

If $f^{\prime}(x)=k(\cos x-\sin x)$,$f^{\prime}(0)=3$,and $f\left(\frac{\pi}{2}\right)=15$,then $f(x)=$

Find the derivative of the function $f(x)=2x^{2}+3x-5$ at $x=-1$. Also,prove that $f^{\prime}(0)+3f^{\prime}(-1)=0$.

Differentiate the function with respect to $x$: $\sin(x^{2}+5)$.

Let $f:R \to R$ be such that $f(1) = 3$ and $f'(1) = 6$. Then $\lim_{x \to 0} \left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}$ equals

If $y = x \left[ \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) + \sin x \right] + \frac{1}{2\sqrt{x}}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module