$f^{-1}$ (বিপরীত ফাংশন) এর ব্যবকলন

Derivative of an Inverse Function

We know how to take lots of derivatives by now. But we're still missing some fundamental derivatives of inverse functions, like $\arcsin x$ or $\log x$. With the tools we have, let's see if we can come up with a way to express the derivative of an inverse function.

What formula relates a function and its inverse?

We know that, if $f^{-1}(x)$ denotes the inverse of $f(x)$, then $f(f^{-1}(x)) = x$.

Let's differentiate both sides.

How do we take the derivative of the left side?

We apply the chain rule! The derivative of the left hand side is $f'(f^{-1}(x))\cdot \left[f^{-1}(x)\right]'$ (intuitively, think about this as the "product of the rates of change of $f$ and $f^{-1}$"), by the chain rule. The derivative of the right hand side, $x$, is just $1$. So we have $$f'(f^{-1}(x)) \cdot \left[f^{-1}(x)\right]' = 1,$$ or, solving for $\left[f^{-1}(x)\right]'$, $$\left[f^{-1}(x)\right]' = \frac{1}{f'(f^{-1}(x))}.$$

So we have developed an explicit formula for the derivative of an inverse function that takes the derivative and inverse of the original function. Cool, huh?

Caveat/food for thought

Maybe you're uncomfortable with some of the steps above. For example,

It turns out this isn't a big deal, since we can basically prove that $f^{-1}$ is differentiable under some natural/intuitive conditions (in other words, there's nothing really surprising going on). If you're interested, we'll cover the proof in another post someday. But it's most important that you understand the intuition behind the chain rule above:

The rates of change of $f$ (more precisely, at the point $f^{-1}(t)$) and $f^{-1}$ (at the point $t$) should multiply to the rate of change of their composition $f(f^{-1}(x)) = x$ (at the point $t$), which is just $1$.

Let $y=f(x)$, so that $$\frac{\text dy}{\text dx}=f'(x)$$

Then $$\frac{\text dx}{\text dy}=\frac1{f'(x)}$$

Now, substitute $x=f^{-1}(x)$: $$\frac{\text df^{-1}(x)}{\text df(f^{-1}(x))}=\frac{\text df^{-1}(x)}{\text dx}=\frac1{f'(f^{-1}(x))}$$

Alternatively, since $f^{-1}(f(x))=x$, by differentiating both sides we obtain $$\frac{\text df^{-1}(f(x))}{\text df(x)}f'(x)=1$$ Again, letting $x=f^{-1}(x)$, we obtain the same result: $$\frac{\text df^{-1}(x)}{\text dx}=\frac1{f'(f^{-1}(x))}$$

Let $y=f^{-1}(x)$. Then, $f(y)=x$.

Differentiating implicitly, $$\frac d{dx}(f(y)) = 1$$ $$f'(y)\frac{dy}{dx} = 1$$ $$\frac{dy}{dx} = \frac1{f'(y)}$$

Substituting $y=f^{-1}(x)$, we get $$\frac{d}{dx}(f^{-1}(x)) = \frac1{f'(f^{-1}(x))}$$