This is a very useful rule and the most difficult to remember. It is used to find the formula for the derivative of a functional expression expressed as a quotient (f(x)=g(x)/h(x)).
To find the derivative, f'(x), start with the expression in the denominator, multiply by the derivative of the numerator, then subtract the product of the numerator and the derivative of the denominator. Then divide everything above by the denominator squared. Not so easy to remember. So, the following should help.
Let di-low represent the derivative of the denominator.
Let di-high represent the derivative of the numerator.
The rule can be remembered easily the following way:
Low di-high minus High di-low all divided by Low squared.
In symbols, it looks like this: f'(x) = [h(x)g'(x)-g(x)h'(x)]/[h(x)]2
Using a prime to designate a derivative, we have the following, more concise form for an expression y=u/v, where u & v are functions of x.
y'=(vu'-uv')/v2
Example:
Let f(x)=(x2 -2x+5)/(3x3 +5x-7) gives f'(x)=[(3x3 +5x-7)(2x-2)-(x2 -2x+5)(9x2 +5)]/(3x +5x-7)2
In prime notation, y=f(x), u=x2 -2x+5, v=3x3 +5x-7 gives y'=the same result as above.
For functions with a constant numerator, the quotient rule is not necessary. In those cases, it is easier to bring the variable denominator “upstairs” (the power on the expression will change sign)
and use the power rule. The following is an example.
If y=20/(x2 -2x+5)2 .You could use the quotient rule but you are making life difficult for yourself.
Instead, bring the denominator upstairs (power will change to -2) to get, y=20(x2 -2x+5)-2 .
Then use the power rule to get the derivative. We get, y'=-40(x2 -2x+5)-3 (2x-2).
Actually, you can do this with all quotients, however, the quotient rule has the advantage when both top & bottom are variable to give the answer over a single denominator. Whereas, the product rule must also be used in conjunction with the power rule the other way resulting in a two term expression. To do the first example this way, express f(x)=(x2 -2x+5)(3x3 +5x-7)-1