Question

Explain $\frac{u}{v}$ rule of differentiation.

Answer 1

Step 1: Necessary conditions

Let $f\left(x\right)$ be a function and $f\left(x\right)$ is ratio of two functions $u\left(x\right)$ and $v\left(x\right)$,

i.e., $f\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}$

where $u$ and $v$ both are differentiable functions and $v\left(x\right)\ne 0$

Step 2: $\frac{u}{v}$ Rule of Differentiation.

$f\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}$

Derivative of $f\left(x\right)$ w.r.t. $x$ is given by,

$f\text{'}\left(x\right)=\frac{u\text{'}\left(x\right)v\left(x\right)-u\left(x\right)v\text{'}\left(x\right)}{v{\left(x\right)}^2}$

Step 3: Proof of the rule.

Definition of Derivative is,

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

$f\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}$. So, $f(x+h)=\frac{u(x+h)}{v(x+h)}$

Now apply the definition of derivative;

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{u(x+h)}{v(x+h)}}-{\displaystyle \frac{u\left(x\right)}{v\left(x\right)}}}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{u(x+h)v\left(x\right)-u\left(x\right)v(x+h)}{v(x+h)v\left(x\right)}}}{h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle u(x+h)v\left(x\right)-u\left(x\right)v(x+h)}}{v(x+h)v\left(x\right)h} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{u(x+h)v\left(x\right)-u\left(x\right)v(x+h)}{h}.\underset{h\to 0}{\lim }\frac{1}{v(x+h)v\left(x\right)}\left[\underset{x\to a}{\lim }\right(u\left(x\right).v\left(x\right))=\underset{x\to a}{\lim }(u\left(x\right)).\underset{x\to a}{\lim }(v\left(x\right)\left)\right] \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{u(x+h)v\left(x\right)-u\left(x\right)v\left(x\right)+u\left(x\right)v\left(x\right)-u\left(x\right)v(x+h)}{h}.\frac{1}{v(x+0)v\left(x\right)}[\mathrm{putting}\mathrm{h}=0\mathrm{in}\mathrm{the}2\mathrm{nd}\mathrm{limit}] \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{v\left(x\right)\left(u\right(x+h)-u(x\left)\right)-u\left(x\right)\left(v\right(x+h)-v(x\left)\right)}{h}.\frac{1}{v{\left(x\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\{\frac{v\left(x\right)\left(u\right(x+h)-u(x\left)\right)}{h}-\frac{u\left(x\right)\left(v\right(x+h)-v(x\left)\right)}{h}\}.\frac{1}{v{\left(x\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\{\underset{h\to 0}{\lim }\frac{v\left(x\right)\left(u\right(x+h)-u(x\left)\right)}{h}-\underset{h\to 0}{\lim }\frac{u\left(x\right)\left(v\right(x+h)-v(x\left)\right)}{h}\}.\frac{1}{v{\left(x\right)}^2}\left[\underset{x\to a}{\lim }\right(f\left(x\right)-g\left(x\right))=\underset{x\to a}{\lim }f(x)-\underset{x\to a}{\lim }g(x\left)\right] \\ \Rightarrow f\text{'}\left(x\right)=\left\{v\right(x).\underset{h\to 0}{\lim }\frac{u(x+h)-u\left(x\right)}{h}-u(x).\underset{h\to 0}{\lim }\frac{v(x+h)-v\left(x\right)}{h}\}.\frac{1}{v{\left(x\right)}^2}[\because ,v(x),u(x)\mathrm{are}\mathrm{constants}] \\ \Rightarrow f\text{'}\left(x\right)=\left\{v\right(x).u\text{'}(x)-u(x).v\text{'}(x\left)\right\}.\frac{1}{v{\left(x\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{u\text{'}\left(x\right)v\left(x\right)-u\left(x\right)v\text{'}\left(x\right)}{v{\left(x\right)}^2} \end{gathered}$$

Hence, the proof is complete.