Quotient rule
There rule similar to the product rule for quotients. To prove it, we go to the definition of the derivative:
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{\displaystyle {\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)
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{\displaystyle {\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
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{\displaystyle (4x-2)/(x^{2}+1)}
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{\displaystyle {\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
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{\displaystyle {\frac {d}{dx}}\left((a+b)\times (c+d)\right)={\frac {d}{dx}}\left(ac+ad+bc+bd\right)}
![{\displaystyle {\frac {d}{dx}}\left[{f(x) \over g(x)}\right]={\frac {g(x)f'(x)-f(x)g'(x)}{g(x)^{2}}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/340b04b8b03645f89015a33c6cc8e0785037a7f6)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/480b0cabadf2fd8512f4c159004ffc2fa904bfed)
