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