$1/f$ এর অন্তরজ

This explanation derives a formula for the derivative of $\frac{1}{f(x)}$ using a trick with the product rule---with a caveat mentioned at the end. It is also possible to derive this using the limit definition or the chain rule.

Ok, great!

The product rule gives that $$\left(f(x)\cdot \frac{1}{f(x)}\right)' = f'(x)\cdot\frac{1}{f(x)} + \left(\frac{1}{f(x)}\right)'\cdot f(x)$$

By above, we can simply the equation from earlier to $$0 = f'(x)\cdot\frac{1}{f(x)} + \left(\frac{1}{f(x)}\right)'\cdot f(x)$$ and we solve for $\left(\frac{1}{f(x)}\right)'$ to get $$\left(\frac{1}{f(x)}\right)' = \frac{-f'(x)}{f(x)\cdot f(x)}$$

There's a caveat here: In using the product rule, we must first assume that $f(x)$ and $\frac{1}{f(x)}$ are differentiable.

In fact, it's true that if $f(x)$ is differentiable and nonzero at a point $a$, then $\frac{1}{f(x)}$ is differentiable at $a$ as well, with derivative $\frac{-f'(a)}{f(a)^2}$. (This can be proven using the limit definition and a change of variables.)

Nonetheless, the product rule trick above is a very nice way to "remember" what the derivative of $\frac{1}{f(x)}$ should be if it exists. Once we've proven that it indeed does have to exist, there's no need to worry about it again.

We can derive a formula for the derivative of a reciprocal with the Quotient Rule:

$\frac{d}{dx}\left(\frac{1}{f(x)}\right) = \frac{f(x)\frac{d}{dx}(1) - 1\frac{d}{dx}f(x)}{f(x)^2}=$

$\frac{f(x)(0) - f'(x)}{f(x)^2} = \frac{-f'(x)}{f(x)^2}$

So obviously the function in the initial denominator must be differentiable for this formula to be usable.

Example: Find the derivative of $\sec(x)$ with this method.

Solution:

$\frac{d}{dx}(\sec(x)) = \frac{d}{dx}\left(\frac{1}{\cos(x)}\right)$

Now apply the formula with $f(x) = \cos(x)$:

$\frac{-\frac{d}{dx}\cos(x)}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)}$

This derivative is more commonly written in the form

$\sec(x)\tan(x)$

One nice way of deriving the product rule for the reciprocal of a function $f$ is to directly use the limit definition of a derivative, and then use some easy manipulations: $$\frac{d}{dx}\left(\frac{1}{f(x)}\right) = \lim_{h\rightarrow 0}\frac{\frac{1}{f(x+h)} - \frac{1}{f(x)}}{h}$$ $$\frac{d}{dx}\left(\frac{1}{f(x)}\right) = \lim_{h\rightarrow 0}\frac{f(x) - f(x+h)}{hf(x)f(x+h)}$$ $$\frac{d}{dx}\left(\frac{1}{f(x)}\right) = \lim_{h\rightarrow 0}\frac{f(x) - f(x+h)}{h} \cdot \lim_{h\rightarrow 0}\frac{1}{f(x)f(x+h)}$$ Now, by the definition of the derivative, the first limit (assuming it exists) equals $-f'(x)$. Also, as $h$ approaches $0$, $f(x+h)$ approaches $f(x)$, so the second limit (assuming it exists) must equal $\frac{1}{f(x)^2}.$

Now, it follows that: $$\frac{d}{dx}\left(\frac{1}{f(x)}\right) = -\frac{f'(x)}{f(x)^2}$$ for all places where $f(x)$ is differentiable and nonzero. This is the derivative of the reciprocal of a function.