Q. 90 - Calculus

Question

The following reciprocal rules tells us hoe to differentiate the reciprocal of a function

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

Prove this using

a) definition of the derivative

b) by using the quotient rule

Step-by-Step Solution

Answer

We prove the reciprocal rule using definition of derivative and quotient rule

Step 1: Given information

We are given the reciprocal rule as $\frac{d}{dx}\left(\frac{1}{f\left(x\right)}\right)=-\frac{1}{{\left[f\right(x\left)\right]}^2}$

Step 2: Prove using the definition

We have

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{f(x+h)}}-{\displaystyle \frac{1}{f\left(x\right)}}}{h} \\ \underset{h\to 0}{\lim }\frac{f\left(x\right)-f(x+h)}{f\left(x\right)f(x+h)h} \\ -\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{f\left(x\right)\left(f\right(x+h\left)\right)h} \\ =-\frac{f\text{'}\left(x\right)}{{\left[f\right(x\left)\right]}^2} \end{gathered}$$

Step 3: Prove using product rule

We have

$$\begin{gathered} \frac{d}{dx}\left[f\right(x\left)g\right(x\left)\right]=f\left(x\right)g\text{'}\left(x\right)+f\text{'}\left(x\right)g\left(x\right) \\ \frac{d}{dx}[\frac{1}{f\left(x\right)}·1]=\frac{1}{f\left(x\right)}·\left(0\right)+[-\frac{f\text{'}\left(x\right)}{f{\left(x\right)}^2}] \\ \frac{d}{dx}\left[\frac{1}{f\left(x\right)}\right]=-\frac{f\text{'}\left(x\right)}{f{\left(x\right)}^2} \end{gathered}$$