If f: R rightarrow R is an even function havi | vedclass

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

Question

(B) Given that $f(x)$ is an even function,so $f(-x) = f(x)$.By differentiating both sides with respect to $x$ using the chain rule:$\frac{d}{dx}[f(-x)

Vedclass · Accepted Answer

$f^{\prime}$

Vedclass · Answer

(B) Given that $f(x)$ is an even function,so $f(-x) = f(x)$.By differentiating both sides with respect to $x$ using the chain rule:$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$$-f^{\prime}(-x) = f^{\prime}(x)$,which implies $f^{\prime}(-x) = -f^{\prime}(x)$.Thus,the first derivative $f^{\prime}(x)$ is an odd function.Now,differentiating again:$\frac{d}{dx}[f^{\prime}(-x)] = \frac{d}{dx}[-f^{\prime}(x)]$$-f^{\prime \prime}(-x) = -f^{\prime \prime}(x)$,which implies $f^{\prime \prime}(-x) = f^{\prime \prime}(x)$.Thus,the second derivative $f^{\prime \prime}(x)$ is an even function.Similarly,the third derivative $f^{\prime \prime \prime}(x)$ will be an odd function.Therefore,$f^{\prime}(x)$ is an odd function. Comparing with the options,option $B$ is correct.

If $f: R \rightarrow R$ is an even function having derivatives of all orders,then which of the following is an odd function?

Similar Questions

Let $f(x)$ be a differentiable function for all $x \in R$ and $f(x+y)=f(x)+f(y)-3xy$. If $\lim _{h \rightarrow 0} \frac{f(h)}{h}=7$,then $f^{\prime}(x)=$

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x-2}{x^2-3x+2} & \text{if } x \in R - \{1, 2\} \\ 2 & \text{if } x = 1 \\ 1 & \text{if } x = 2 \end{cases}$,then find $\lim_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}$.

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

The derivative of $F[f\{ \phi (x)\} ]$ with respect to $x$ is:

If $f(x) = \frac{x}{1+x}$ and $g(x) = f(f(x))$,then $g^{\prime}(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module