Product Rule. If f f f g g g ( f ( x ) g ( x ) ) ′ = f ( x ) g ′ ( x ) + f ′ ( x ) g ( x ) {\left({f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\right)}'={f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}+{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}} ( f ( x ) g ( x ) ) ′ = f ( x ) g ′ ( x ) + f ′ ( x ) g ( x )
Proof.
By definition
( f ( x ) g ( x ) ) = lim h → 0 f ( x + h ) g ( x + h ) − f ( x ) g ( x ) h = {\left({f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\right)}=\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}{g{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}{g{{\left({x}\right)}}}}}{{h}}= ( f ( x ) g ( x ) ) = lim h → 0 h f ( x + h ) g ( x + h ) − f ( x ) g ( x ) =
= lim h → 0 f ( x + h ) g ( x + h ) − f ( x + h ) g ( x ) + f ( x + h ) g ( x ) − f ( x ) g ( x ) h = =\lim_{{{h}\to{0}}}\frac{{{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}}= = lim h → 0 h f ( x + h ) g ( x + h ) − f ( x + h ) g ( x ) + f ( x + h ) g ( x ) − f ( x ) g ( x ) =
= lim h → 0 f ( x + h ) g ( x + h ) − f ( x + h ) g ( x ) h + lim h → 0 f ( x + h ) g ( x ) − f ( x ) g ( x ) h = =\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}{g{{\left({x}+{h}\right)}}}-{f{{\left({x}+{h}\right)}}}{g{{\left({x}\right)}}}}}{{h}}+\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}{g{{\left({x}\right)}}}-{f{{\left({x}\right)}}}{g{{\left({x}\right)}}}}}{{h}}= = lim h → 0 h f ( x + h ) g ( x + h ) − f ( x + h ) g ( x ) + lim h → 0 h f ( x + h ) g ( x ) − f ( x ) g ( x ) =
= lim h → 0 ( f ( x + h ) g ( x + h ) − g ( x ) h ) + lim h → 0 ( g ( x ) f ( x + h ) − f ( x ) h ) = =\lim_{{{h}\to{0}}}{\left({f{{\left({x}+{h}\right)}}}\frac{{{g{{\left({x}+{h}\right)}}}-{g{{\left({x}\right)}}}}}{{h}}\right)}+\lim_{{{h}\to{0}}}{\left({g{{\left({x}\right)}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}\right)}= = lim h → 0 ( f ( x + h ) h g ( x + h ) − g ( x ) ) + lim h → 0 ( g ( x ) h f ( x + h ) − f ( x ) ) =
lim h → 0 f ( x + h ) lim h → 0 g ( x + h ) − g ( x ) h + lim h → 0 g ( x ) lim h → 0 f ( x + h ) − f ( x ) h = \lim_{{{h}\to{0}}}{f{{\left({x}+{h}\right)}}}\lim_{{{h}\to{0}}}\frac{{{g{{\left({x}+{h}\right)}}}-{g{{\left({x}\right)}}}}}{{h}}+\lim_{{{h}\to{0}}}{g{{\left({x}\right)}}}\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}= lim h → 0 f ( x + h ) lim h → 0 h g ( x + h ) − g ( x ) + lim h → 0 g ( x ) lim h → 0 h f ( x + h ) − f ( x ) =
= f ( x + 0 ) g ′ ( x ) + g ( x ) f ′ ( x ) = f ( x ) g ′ ( x ) + f ′ ( x ) g ( x ) ={f{{\left({x}+{0}\right)}}}{g{'}}{\left({x}\right)}+{g{{\left({x}\right)}}}{f{'}}{\left({x}\right)}={f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}+{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}} = f ( x + 0 ) g ′ ( x ) + g ( x ) f ′ ( x ) = f ( x ) g ′ ( x ) + f ′ ( x ) g ( x )
Example 1 .f ( x ) = x e x {f{{\left({x}\right)}}}={x}{{e}}^{{x}} f ( x ) = x e x
f ( x ) ′ = ( x e x ) ′ = x ( e x ) ′ + ( x ′ ) e x = x e x + 1 ⋅ e x = e x ( x + 1 ) {f{{\left({x}\right)}}}'={\left({x}{{e}}^{{x}}\right)}'={x}{\left({{e}}^{{x}}\right)}'+{\left({x}'\right)}{{e}}^{{x}}={x}{{e}}^{{x}}+{1}\cdot{{e}}^{{x}}={{e}}^{{x}}{\left({x}+{1}\right)} f ( x ) ′ = ( x e x ) ′ = x ( e x ) ′ + ( x ′ ) e x = x e x + 1 ⋅ e x = e x ( x + 1 )
Example2 .f ( t ) = t 2 t {f{{\left({t}\right)}}}={{t}}^{{2}}\sqrt{{{t}}} f ( t ) = t 2 t
f ′ ( t ) = ( t 2 t ) ′ = t 2 ( t ) ′ + ( t 2 ) ′ t = t 2 1 2 t + 2 t t = 5 2 t t {f{'}}{\left({t}\right)}={\left({{t}}^{{2}}\sqrt{{{t}}}\right)}'={{t}}^{{2}}{\left(\sqrt{{{t}}}\right)}'+{\left({{t}}^{{2}}\right)}'\sqrt{{{t}}}={{t}}^{{2}}\frac{{1}}{{{2}\sqrt{{{t}}}}}+{2}{t}\sqrt{{{t}}}=\frac{{5}}{{2}}{t}\sqrt{{{t}}} f ′ ( t ) = ( t 2 t ) ′ = t 2 ( t ) ′ + ( t 2 ) ′ t = t 2 2 t 1 + 2 t t = 2 5 t t
Note, however that Example 2 can be solved without product rule. Since t 2 t = t 2 t 1 2 = t 5 2 {{t}}^{{2}}\sqrt{{{t}}}={{t}}^{{2}}{{t}}^{{\frac{{1}}{{2}}}}={{t}}^{{\frac{{5}}{{2}}}} t 2 t = t 2 t 2 1 = t 2 5 f ′ ( t ) = 5 2 t 5 2 − 1 = 5 2 t 3 2 {f{'}}{\left({t}\right)}=\frac{{5}}{{2}}{{t}}^{{\frac{{5}}{{2}}-{1}}}=\frac{{5}}{{2}}{{t}}^{{\frac{{3}}{{2}}}} f ′ ( t ) = 2 5 t 2 5 − 1 = 2 5 t 2 3
Example 2 showed that first you need to look at the function and make sure that product rule is the only way to find derivative (like in Example 1).
Product rule can be extended on case of more than two functions.
Let's see how this can be done for three functions:
( u v w ) ′ = ( u v ) ′ w + u v w ′ = ( u ′ v + u v ′ ) w + u v w ′ = u ′ v w + u v ′ w + u v w ′ {\left({u}{v}{w}\right)}'={\left({u}{v}\right)}'{w}+{u}{v}{w}'={\left({u}'{v}+{u}{v}'\right)}{w}+{u}{v}{w}'={u}'{v}{w}+{u}{v}'{w}+{u}{v}{w}' ( u v w ) ′ = ( u v ) ′ w + u v w ′ = ( u ′ v + u v ′ ) w + u v w ′ = u ′ v w + u v ′ w + u v w ′
Similarly can be proven case for four functions:
( u v w z ) ′ = u ′ v w z + u v ′ w z + u v w ′ z + u v w z ′ {\left({u}{v}{w}{z}\right)}'={u}'{v}{w}{z}+{u}{v}'{w}{z}+{u}{v}{w}'{z}+{u}{v}{w}{z}' ( u v w z ) ′ = u ′ v w z + u v ′ w z + u v w ′ z + u v w z ′
Do you see pattern? In general if we have n {n} n n {n} n
Example 3 .y = x 2 sin ( x ) e x {y}={{x}}^{{2}}{\sin{{\left({x}\right)}}}{{e}}^{{x}} y = x 2 sin ( x ) e x
y ′ = ( x 2 sin ( x ) e x ) ′ = ( x 2 ) ′ sin ( x ) e x + x 2 ( sin ( x ) ) ′ e x + x 2 sin ( x ) ( e x ) ′ = {y}'={\left({{x}}^{{2}}{\sin{{\left({x}\right)}}}{{e}}^{{x}}\right)}'={\left({{x}}^{{2}}\right)}'{\sin{{\left({x}\right)}}}{{e}}^{{x}}+{{x}}^{{2}}{\left({\sin{{\left({x}\right)}}}\right)}'{{e}}^{{x}}+{{x}}^{{2}}{\sin{{\left({x}\right)}}}{\left({{e}}^{{x}}\right)}'= y ′ = ( x 2 sin ( x ) e x ) ′ = ( x 2 ) ′ sin ( x ) e x + x 2 ( sin ( x ) ) ′ e x + x 2 sin ( x ) ( e x ) ′ =
= 2 x sin ( x ) e x + x 2 cos ( x ) e x + x 2 sin ( x ) e x = x e x ( 2 sin ( x ) + x cos ( x ) + x sin ( x ) ) ={2}{x}{\sin{{\left({x}\right)}}}{{e}}^{{x}}+{{x}}^{{2}}{\cos{{\left({x}\right)}}}{{e}}^{{x}}+{{x}}^{{2}}{\sin{{\left({x}\right)}}}{{e}}^{{x}}={x}{{e}}^{{x}}{\left({2}{\sin{{\left({x}\right)}}}+{x}{\cos{{\left({x}\right)}}}+{x}{\sin{{\left({x}\right)}}}\right)} = 2 x sin ( x ) e x + x 2 cos ( x ) e x + x 2 sin ( x ) e x = x e x ( 2 sin ( x ) + x cos ( x ) + x sin ( x ) )
Example 4 .( x 2 sin ( x ) ln ( x ) e x ) ′ {\left({{x}}^{{2}}{\sin{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{{e}}^{{x}}\right)}' ( x 2 sin ( x ) ln ( x ) e x ) ′
( x 2 sin ( x ) ln ( x ) e x ) ′ = {\left({{x}}^{{2}}{\sin{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{{e}}^{{x}}\right)}'= ( x 2 sin ( x ) ln ( x ) e x ) ′ =
= ( x 2 ) ′ sin ( x ) ln ( x ) e x + x 2 ( sin ( x ) ) ′ ln ( x ) e x + x 2 sin ( x ) ( ln ( x ) ) ′ e x + x 2 sin ( x ) ln ( x ) ( e x ) ′ = ={\left({{x}}^{{2}}\right)}'{\sin{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{{e}}^{{x}}+{{x}}^{{2}}{\left({\sin{{\left({x}\right)}}}\right)}'{\ln{{\left({x}\right)}}}{{e}}^{{x}}+{{x}}^{{2}}{\sin{{\left({x}\right)}}}{\left({\ln{{\left({x}\right)}}}\right)}'{{e}}^{{x}}+{{x}}^{{2}}{\sin{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{\left({{e}}^{{x}}\right)}'= = ( x 2 ) ′ sin ( x ) ln ( x ) e x + x 2 ( sin ( x ) ) ′ ln ( x ) e x + x 2 sin ( x ) ( ln ( x ) ) ′ e x + x 2 sin ( x ) ln ( x ) ( e x ) ′ =
= 2 x sin ( x ) ln ( x ) x 2 cos ( x ) ln ( x ) e x + x sin ( x ) e x + x 2 sin ( x ) ln ( x ) e x ={2}{x}{\sin{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{{x}}^{{2}}{\cos{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{{e}}^{{x}}+{x}{\sin{{\left({x}\right)}}}{{e}}^{{x}}+{{x}}^{{2}}{\sin{{\left({x}\right)}}}{\ln{{\left({x}\right)}}}{{e}}^{{x}} = 2 x sin ( x ) ln ( x ) x 2 cos ( x ) ln ( x ) e x + x sin ( x ) e x + x 2 sin ( x ) ln ( x ) e x
