Question

Left hand derivative and right hand derivative of a function $f\left(x\right)$ at a point $x=a$ are defined as

$f^{\prime }\left(a^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{f\left(a\right)-f(a-h)}{h}=\underset{h\to 0^{-}}{lim}\frac{f\left(a\right)-f(a-h)}{h}=\underset{x\to a^{+}}{lim}\frac{f\left(a\right)-f\left(x\right)}{a-x}$ respectively

Let $f$ be a twice differentiable function. We also know that derivative of an even function is odd function and derivative of an odd function is even function.

If $f$ is odd, which of the following is Left-hand derivative of $f$ at $x=a$

• A. $\underset{h\to 0^{-}}{lim}\frac{f(a-h)-f\left(a\right)}{-h}$
• B. $\underset{h\to 0^{-}}{lim}\frac{f(a-h)-f\left(a\right)}{h}$
• C. $\underset{h\to 0^{+}}{lim}\frac{f\left(a\right)+-f(a-h)}{-h}$
• D. $\underset{h\to 0^{-}}{lim}\frac{f(-a)-f(-a-h)}{-h}$

Answer 1

The correct option is A $\underset{h\to 0^{-}}{lim}\frac{f(a-h)-f\left(a\right)}{-h}$

According to the first principle of deriative,

A derivative is simply a measure of the rate of the change

whose $y=f\left(x\right)$ with respect to its variable x

GIven : $f^1\left(a\_\right)={lim}_{h\to 0^{+}}\frac{f\left(a\right)-f(a-h)}{h}$

$={lim}_{h\to 0^{-}}\frac{f\left(a\right)-f(a-h)}{h}$

$={lim}_{x\to 0^{+}}\frac{f\left(a\right)-f\left(x\right)}{a-x}$

W.K.T differntiation of an even function is odd and vice versa.

T.F.T of x is odd

then left hand derivative of f at $x=0$

$={lim}_{h\to 0^{-}}\frac{f(a-h)-f\left(a\right)}{-h}$

since $x$ is odd and $x=a$ ,the derivative should be an even function

which should be of the form

$f^1\left(x\right)={lim}_{h\to 0^{-}}\frac{f(x-h)-f\left(x\right)}{h}substitutex=a$.