Math 22 Product Rule and Quotient Rule

The Product Rule

 The derivative of the product of two differentiable functions is equal to the first function times 
 the derivative of the second plus the second function times the derivative of the first.
 ${\frac {d}{dx}}[f(x)\cdot g(x)]=f'(x)g(x)+f(x)g'(x)$

Example: Find derivative of

1) $f(x)=(x+1)(x^{2}+3)$

Solution:
$f'(x)={\frac {d}{dx}}[(x+1)](x^{2}+3)+(x+1){\frac {d}{dx}}[(x^{2}+3)]$
$=(1)(x^{2}+3)+(x+1)(2x)=x^{2}+3+2x^{2}+2x=3x^{2}+2x+3$

2) $f(x)=(4x+3x^{2})(6-3x)$

Solution:
$f'(x)={\frac {d}{dx}}[(4x+3x^{2})](6-3x)+(4x+3x^{2}){\frac {d}{dx}}[(6-3x)]$
$=(4+6x)(6-3x)+(4x+3x^{2})(-3)=24+36x-12x-18x^{2}-12x-9x^{2}=-27x^{2}+12x+24$

3) $f(x)=(e^{2}+x^{2})(4x+5)$

Solution:
$f'(x)={\frac {d}{dx}}[(e^{2}+x^{2})](4x+5)+(e^{2}+x^{2}){\frac {d}{dx}}[(4x+5)]$
$=(2x)(4x+5)+(e^{2}+x^{2})(4)=8x^{2}+10x+4e^{2}+8x^{2}=-16x^{2}+10x+4e^{2}$

The Quotient Rule

 The derivative of the quotient of two differentiable functions is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the 
 denominator.
 ${\frac {d}{dx}}[{\frac {f(x)}{g(x)}}]={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}$

Example: Find derivative of

1) $f(x)={\frac {x}{x-5}}$

Solution:
$f'(x)={\frac {{\frac {d}{dx}}[x](x-5)-(x){\frac {d}{dx}}[(x-5)]}{(x-5)^{2}}}$
$={\frac {(1)(x-5)-x(1)}{(x-5)^{2}}}={\frac {-5}{(x-5)^{2}}}$

2) $f(x)={\frac {x^{2}}{x+3}}$

Solution:
$f'(x)={\frac {{\frac {d}{dx}}[x^{2}](x+3)-(x^{2}){\frac {d}{dx}}[(x+3)]}{(x+3)^{2}}}$
$={\frac {(2x)(x+3)-(x^{2})(1)}{(x+3)^{2}}}={\frac {2x^{2}+6x-x^{2}}{(x+3)^{2}}}={\frac {x^{2}+6x}{(x+3)^{2}}}$

3) $f(x)={\frac {x^{2}+6x+5}{2x-1}}$

Solution:
$f'(x)={\frac {{\frac {d}{dx}}[x^{2}+6x+5](2x-1)-(x^{2}+6x+5){\frac {d}{dx}}[(2x-1)]}{(2x-1)^{2}}}$
$={\frac {(2x+6)(2x-1)-(x^{2}+6x+5)(2)}{(2x-1)^{2}}}={\frac {4x^{2}+12x-2x-6-2x^{2}-12x-10}{(2x-1)^{2}}}={\frac {2x^{2}-2x-16}{(2x-1)^{2}}}$

Return to Topics Page

This page were made by Tri Phan