Q. 89 - Calculus

Question

use the definition of the derivative to prove the quotient rule

Step-by-Step Solution

Answer

We use the definition of derivative to prove the quotient rule

Step 1: Given information

We are given a function of the form $\frac{f\left(x\right)}{g\left(x\right)}$

Step 2: use the definition to find the derivative

We get,

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