# Calculus/Quotient Rule

 ← Product Rule Calculus Derivatives of Trigonometric Functions → Quotient Rule

Quotient rule

There rule similar to the product rule for quotients. To prove it, we go to the definition of the derivative:

{\begin{aligned}{\frac {d}{dx}}{\frac {f(x)}{g(x)}}&=\lim _{h\to 0}{\frac {{\frac {f(x+h)}{g(x+h)}}-{\frac {f(x)}{g(x)}}}{h}}\\&=\lim _{h\to 0}{\frac {f(x+h)g(x)-f(x)g(x+h)}{hg(x)g(x+h)}}\\&=\lim _{h\to 0}{\frac {f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{hg(x)g(x+h)}}\\&=\lim _{h\to 0}{\frac {g(x){\frac {f(x+h)-f(x)}{h}}-f(x){\frac {g(x+h)-g(x)}{h}}}{g(x)g(x+h)}}\\&={\frac {g(x)f'(x)-f(x)g'(x)}{g(x)^{2}}}\end{aligned}} This leads us to the so-called "quotient rule":

 Derivatives of quotients (Quotient Rule)${\frac {d}{dx}}\left[{f(x) \over g(x)}\right]={\frac {g(x)f'(x)-f(x)g'(x)}{g(x)^{2}}}\,\!$ Which some people remember with the mnemonic "low D-high minus high D-low (over) square the low and away we go!"

## Examples

The derivative of $(4x-2)/(x^{2}+1)$  is:

{\begin{aligned}{\frac {d}{dx}}\left[{\frac {(4x-2)}{x^{2}+1}}\right]&={\frac {(x^{2}+1)(4)-(4x-2)(2x)}{(x^{2}+1)^{2}}}\\&={\frac {(4x^{2}+4)-(8x^{2}-4x)}{(x^{2}+1)^{2}}}\\&={\frac {-4x^{2}+4x+4}{(x^{2}+1)^{2}}}\end{aligned}}

Remember: the derivative of a product/quotient is not the product/quotient of the derivatives. (That is, differentiation does not distribute over multiplication or division.) However one can distribute before taking the derivative. That is ${\frac {d}{dx}}\left((a+b)\times (c+d)\right)={\frac {d}{dx}}\left(ac+ad+bc+bd\right)$