Theorem 4. If u u u v v v x x x y = u v y=\dfrac uv y = v u v ≠ 0 v\ne0 v = 0
d y d x = v d u d x − u d v d x v 2 \dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2} d x d y = v 2 v d x d u − u d x d v
Proof. Given y = u v y=\dfrac uv y = v u x x x δ x \delta x δ x u → u + δ u u\to u+\delta u u → u + δ u v → v + δ v v\to v+\delta v v → v + δ v y → y + δ y y\to y+\delta y y → y + δ y
y + δ y = u + δ u v + δ v y+\delta y=\dfrac{u+\delta u}{v+\delta v} y + δ y = v + δ v u + δ u
δ y = u + δ u v + δ v − u v = v ( u + δ u ) − u ( v + δ v ) v ( v + δ v ) = v δ u − u δ v v ( v + δ v ) \delta y=\dfrac{u+\delta u}{v+\delta v}-\dfrac uv=\dfrac{v(u+\delta u)-u(v+\delta v)}{v(v+\delta v)}=\dfrac{v\,\delta u-u\,\delta v}{v(v+\delta v)} δ y = v + δ v u + δ u − v u = v ( v + δ v ) v ( u + δ u ) − u ( v + δ v ) = v ( v + δ v ) v δ u − u δ v
Since δ x ≠ 0 \delta x\ne0 δ x = 0 δ x \delta x δ x
δ y δ x = v δ u δ x − u δ v δ x v ( v + δ v ) \dfrac{\delta y}{\delta x}=\dfrac{v\dfrac{\delta u}{\delta x}-u\dfrac{\delta v}{\delta x}}{v(v+\delta v)} δ x δ y = v ( v + δ v ) v δ x δ u − u δ x δ v
Taking the limit as δ x → 0 \delta x\to0 δ x → 0 δ v → 0 \delta v\to0 δ v → 0 v v v
lim δ x → 0 δ y δ x = v lim δ x → 0 δ u δ x − u lim δ x → 0 δ v δ x v ( v + lim δ x → 0 δ v ) = v d u d x − u d v d x v 2 \lim_{\delta x\to0}\dfrac{\delta y}{\delta x}=\dfrac{v\lim\limits_{\delta x\to0}\dfrac{\delta u}{\delta x}-u\lim\limits_{\delta x\to0}\dfrac{\delta v}{\delta x}}{v\left(v+\lim\limits_{\delta x\to0}\delta v\right)}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2} lim δ x → 0 δ x δ y = v ( v + δ x → 0 lim δ v ) v δ x → 0 lim δ x δ u − u δ x → 0 lim δ x δ v = v 2 v d x d u − u d x d v
using the differentiability of u u u v v v
d y d x = v d u d x − u d v d x v 2 ■ \dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}\qquad\blacksquare d x d y = v 2 v d x d u − u d x d v ■
Worked examples (combining the sum, product and quotient rules).
Example 1(1): y = x 3 / 2 + log x − cos x y=x^{3/2}+\log x-\cos x y = x 3/2 + log x − cos x Differentiating term by term (sum/difference rule):
d y d x = d d x ( x 3 / 2 ) + d d x ( log x ) − d d x ( cos x ) = 3 2 x 1 / 2 + 1 x − ( − sin x ) = 3 2 x + 1 x + sin x \dfrac{dy}{dx}=\dfrac{d}{dx}\!\left(x^{3/2}\right)+\dfrac{d}{dx}(\log x)-\dfrac{d}{dx}(\cos x)=\dfrac32x^{1/2}+\dfrac1x-(-\sin x)=\dfrac32\sqrt x+\dfrac1x+\sin x d x d y = d x d ( x 3/2 ) + d x d ( log x ) − d x d ( cos x ) = 2 3 x 1/2 + x 1 − ( − sin x ) = 2 3 x + x 1 + sin x
Example 1(2): f ( x ) = x 5 cosec x + x tan x f(x)=x^5\,\text{cosec}\,x+\sqrt x\tan x f ( x ) = x 5 cosec x + x tan x Both terms are products, so the product rule is applied to each:
f ′ ( x ) = [ x 5 ⋅ d d x ( cosec x ) + cosec x ⋅ d d x ( x 5 ) ] + [ x ⋅ d d x ( tan x ) + tan x ⋅ d d x ( x ) ] f'(x)=\Big[x^5\cdot\dfrac{d}{dx}(\text{cosec}\,x)+\text{cosec}\,x\cdot\dfrac{d}{dx}(x^5)\Big]+\Big[\sqrt x\cdot\dfrac{d}{dx}(\tan x)+\tan x\cdot\dfrac{d}{dx}(\sqrt x)\Big] f ′ ( x ) = [ x 5 ⋅ d x d ( cosec x ) + cosec x ⋅ d x d ( x 5 ) ] + [ x ⋅ d x d ( tan x ) + tan x ⋅ d x d ( x ) ]
= x 5 ( − cosec x cot x ) + cosec x ( 5 x 4 ) + x ( sec 2 x ) + tan x ( 1 2 x ) =x^5(-\text{cosec}\,x\cot x)+\text{cosec}\,x\,(5x^4)+\sqrt x\,(\sec^2x)+\tan x\left(\dfrac1{2\sqrt x}\right) = x 5 ( − cosec x cot x ) + cosec x ( 5 x 4 ) + x ( sec 2 x ) + tan x ( 2 x 1 )
= − x 5 cosec x cot x + 5 x 4 cosec x + x sec 2 x + tan x 2 x =-x^5\,\text{cosec}\,x\cot x+5x^4\,\text{cosec}\,x+\sqrt x\sec^2x+\dfrac{\tan x}{2\sqrt x} = − x 5 cosec x cot x + 5 x 4 cosec x + x sec 2 x + 2 x tan x
Example 1(3): y = e x − 5 e x + 5 y=\dfrac{e^x-5}{e^x+5} y = e x + 5 e x − 5 By the quotient rule, with u = e x − 5 , v = e x + 5 u=e^x-5,\ v=e^x+5 u = e x − 5 , v = e x + 5
d y d x = ( e x + 5 ) ⋅ e x − ( e x − 5 ) ⋅ e x ( e x + 5 ) 2 = e 2 x + 5 e x − e 2 x + 5 e x ( e x + 5 ) 2 = 10 e x ( e x + 5 ) 2 \dfrac{dy}{dx}=\dfrac{(e^x+5)\cdot e^x-(e^x-5)\cdot e^x}{(e^x+5)^2}=\dfrac{e^{2x}+5e^x-e^{2x}+5e^x}{(e^x+5)^2}=\dfrac{10e^x}{(e^x+5)^2} d x d y = ( e x + 5 ) 2 ( e x + 5 ) ⋅ e x − ( e x − 5 ) ⋅ e x = ( e x + 5 ) 2 e 2 x + 5 e x − e 2 x + 5 e x = ( e x + 5 ) 2 10 e x
Example 1(4): y = x sin x x + sin x y=\dfrac{x\sin x}{x+\sin x} y = x + sin x x sin x By the quotient rule, with u = x sin x , v = x + sin x u=x\sin x,\ v=x+\sin x u = x sin x , v = x + sin x d u d x = x cos x + sin x \dfrac{du}{dx}=x\cos x+\sin x d x d u = x cos x + sin x d v d x = 1 + cos x \dfrac{dv}{dx}=1+\cos x d x d v = 1 + cos x
d y d x = ( x + sin x ) ( x cos x + sin x ) − x sin x ( 1 + cos x ) ( x + sin x ) 2 \dfrac{dy}{dx}=\dfrac{(x+\sin x)(x\cos x+\sin x)-x\sin x(1+\cos x)}{(x+\sin x)^2} d x d y = ( x + sin x ) 2 ( x + sin x ) ( x cos x + sin x ) − x sin x ( 1 + cos x )
Expanding the numerator: ( x + sin x ) ( x cos x + sin x ) = x 2 cos x + x sin x + x sin x cos x + sin 2 x (x+\sin x)(x\cos x+\sin x)=x^2\cos x+x\sin x+x\sin x\cos x+\sin^2x ( x + sin x ) ( x cos x + sin x ) = x 2 cos x + x sin x + x sin x cos x + sin 2 x x sin x ( 1 + cos x ) = x sin x + x sin x cos x x\sin x(1+\cos x)=x\sin x+x\sin x\cos x x sin x ( 1 + cos x ) = x sin x + x sin x cos x x sin x x\sin x x sin x x sin x cos x x\sin x\cos x x sin x cos x
d y d x = x 2 cos x + sin 2 x ( x + sin x ) 2 \dfrac{dy}{dx}=\dfrac{x^2\cos x+\sin^2x}{(x+\sin x)^2} d x d y = ( x + sin x ) 2 x 2 cos x + sin 2 x