Solveeit Logo
Login

Question

Question: Like \(\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}\)...

Like ddx(uv)=vddxu−uddxvv2\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 ddx(uv)\dfrac{d}{dx}\left( uv \right) .

Explanation

Solution

We have to consider u=f(x),v=g(x)u=f\left( x \right),v=g\left( x \right) and F(x)=uvF\left( x \right)=uv . We will then consider F(x)=f(x)g(x)F\left( x \right)=f\left( x \right)g\left( x \right) and F(x+h)=f(x+h)g(x+h)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′(x)=lim⁡h→0F(x+h)−F(x)h{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(x+h)g(x)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 ddx(uv)\dfrac{d}{dx}\left( uv \right) like the rule ddx(uv)=vddxu−uddxvv2\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(x)...(a)u=f\left( x \right)...\left( a \right) , v=g(x)...(b)v=g\left( x \right)...\left( b \right) and F(x)=uv...(c)F\left( x \right)=uv...\left( c \right) . Let us also consider F(x)=f(x)g(x)...(i)F\left( x \right)=f\left( x \right)g\left( x \right)...\left( i \right) .
Let us replace x with x+hx+h in equation (i).
⇒F(x+h)=f(x+h)g(x+h)...(ii)\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′(x)=lim⁡h→0F(x+h)−F(x)h{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.
⇒F′(x)=lim⁡h→0f(x+h)g(x+h)−f(x)g(x)h\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(x+h)g(x)f\left( x+h \right)g\left( x \right) in the numerator.
⇒F′(x)=lim⁡h→0f(x+h)g(x+h)−f(x+h)g(x)+f(x+h)g(x)−f(x)g(x)h\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.

⇒F′(x)=lim⁡h→0f(x+h)(g(x+h)−g(x))+g(x)(f(x+h)−f(x))h\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
⇒F′(x)=lim⁡h→0[f(x+h)g(x+h)−g(x)h+g(x)f(x+h)−f(x)h]\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 lim⁡x→af(x)g(x)=lim⁡x→af(x)×lim⁡x→ag(x)\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
⇒F′(x)=lim⁡h→0f(x+h)lim⁡h→0g(x+h)−g(x)h+lim⁡h→0g(x)lim⁡h→0f(x+h)−f(x)h\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′(x)=lim⁡h→0f(x+h)−f(x)h{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
⇒F′(x)=lim⁡h→0f(x+h)g′(x)+lim⁡h→0g(x)×f′(x)\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, lim⁡x→ag(y)=g(y)\displaystyle \lim_{x\to a}g\left( y \right)=g\left( y \right) .

& \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{aligned}$$ Let us substitute (a) , (b) and (c) in the above equation. $$\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.