Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Like $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$ , what is the rule for $\dfrac{d}{dx}\left( uv \right)$ .

seo-qna
Last updated date: 12th Jul 2024
Total views: 345.9k
Views today: 5.45k
Answer
VerifiedVerified
345.9k+ views
Hint: We have to consider $u=f\left( x \right),v=g\left( x \right)$ and $F\left( x \right)=uv$ . We will then consider $F\left( x \right)=f\left( x \right)g\left( x \right)$ and $F\left( x+h \right)=f\left( x+h \right)g\left( x+h \right)$ . Then, we will use the definition of derivatives, that is, ${F}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{F\left( x+h \right)-F\left( x \right)}{h}$ . We will substitute the above values in this formula. Then, we have to add and subtract $f\left( x+h \right)g\left( x \right)$ in the numerator. We will then take the common terms outside and apply the limits.

Complete step by step answer:
We have to obtain the rule for $\dfrac{d}{dx}\left( uv \right)$ like the rule $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$ . Let us consider $u=f\left( x \right)...\left( a \right)$ , $v=g\left( x \right)...\left( b \right)$ and $F\left( x \right)=uv...\left( c \right)$ . Let us also consider $F\left( x \right)=f\left( x \right)g\left( x \right)...\left( i \right)$ .
Let us replace x with $x+h$ in equation (i).
$\Rightarrow F\left( x+h \right)=f\left( x+h \right)g\left( x+h \right)...\left( ii \right)$
We know that derivative of a function F(x) is given by
${F}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{F\left( x+h \right)-F\left( x \right)}{h}$
Let us substitute (i) and (ii) in the above equation.
\[\Rightarrow {F}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)g\left( x+h \right)-f\left( x \right)g\left( x \right)}{h}\]
We have to add and subtract $f\left( x+h \right)g\left( x \right)$ in the numerator.
\[\Rightarrow {F}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)g\left( x+h \right)-f\left( x+h \right)g\left( x \right)+f\left( x+h \right)g\left( x \right)-f\left( x \right)g\left( x \right)}{h}\]
Let us take common terms outside.

\[\Rightarrow {F}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)\left( g\left( x+h \right)-g\left( x \right) \right)+g\left( x \right)\left( f\left( x+h \right)-f\left( x \right) \right)}{h}\]
We can rewrite the above equation as
\[\Rightarrow {F}'\left( x \right)=\displaystyle \lim_{h \to 0}\left[ f\left( x+h \right)\dfrac{g\left( x+h \right)-g\left( x \right)}{h}+g\left( x \right)\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \right]\]
We know that $\displaystyle \lim_{x\to a}f\left( x \right)g\left( x \right)=\displaystyle \lim_{x\to a}f\left( x \right)\times \displaystyle \lim_{x\to a}g\left( x \right)$ . Therefore, we can write the above equation as
\[\Rightarrow {F}'\left( x \right)=\displaystyle \lim_{h \to 0}f\left( x+h \right)\displaystyle \lim_{h \to 0}\dfrac{g\left( x+h \right)-g\left( x \right)}{h}+\displaystyle \lim_{h \to 0}g\left( x \right)\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\]
We know that derivative of a function f(x) is given by the formula ${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . Therefore, the above equation becomes
\[\Rightarrow {F}'\left( x \right)=\displaystyle \lim_{h \to 0}f\left( x+h \right){g}'\left( x \right)+\displaystyle \lim_{h \to 0}g\left( x \right)\times {f}'\left( x \right)\]
Let us apply the limits. We know that the limit of a constant is constant, that is, $\displaystyle \lim_{x\to a}g\left( y \right)=g\left( y \right)$ .
\[\begin{align}
  & \Rightarrow {F}'\left( x \right)=f\left( x+0 \right){g}'\left( x \right)+g\left( x \right){f}'\left( x \right) \\
 & \Rightarrow {F}'\left( x \right)=f\left( x \right){g}'\left( x \right)+g\left( x \right){f}'\left( x \right) \\
\end{align}\]
Let us substitute (a) , (b) and (c) in the above equation.
\[\Rightarrow \dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\]

Note: Students must be thorough with the derivative formula in terms of the limits. They have a chance of making error by writing the derivative formula as ${f}'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)+f\left( x \right)}{h}$ . Students must understand the properties of limits and how to apply the limits. We commonly call the rule \[\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\] as product rule.