Derivative : Quotient Rule
Introduction
In calculus, we often encounter functions that are ratios (fractions) of two other functions — for example:
When you need to find the derivative of such a function, you cannot simply divide the derivatives of the numerator and denominator.
That’s where the Quotient Rule comes in.
It gives us a method for differentiating functions written as one function divided by another.
The Quotient Rule Formula
If: $$f(x)=\frac{u(x)}{v(x)}$$
where both uu and vv are differentiable functions,
then the derivative is given by:
example)
$$f(x)=\frac{1}{x+2}$$ then
$$f^\prime(x)=\frac{1^\prime\cdot(x+2)-1\cdot(x+2)^\prime}{(x+2)^2}$$
$$f^\prime(x)=\frac{-1}{(x+2)^2}$$
$$g(x)=\frac{2x+2}{x+5}$$ then
$$g^\prime(x)=\frac{(2x+2)^\prime\cdot(x+5)-(2x+2)\cdot(x+5)^\prime}{(x+5)^2}$$
$$g^\prime(x)=\frac{2\cdot(x+5)-1\cdot(2x+2)}{(x+5)^2}$$
$$g^\prime(x)=\frac{2x+10-2x-2}{(x+5)^2}$$
$$g^\prime(x)=\frac{8}{(x+5)^2}$$
$$h(x)=\frac{(x+4)^2}{(3x+7)^2}=\frac{x^2+8x+16}{9x^2+42x+49}$$ then
$$h^\prime(x)=\frac{(x^2+8x+16)^\prime\cdot(9x^2+42x+49)-(x^2+8x+16)\cdot(9x^2+42x+49)^\prime}{((3x+7)^2)^2}$$
$$h^\prime(x)=\frac{(2x+8)\cdot(9x^2+42x+49)-(x^2+8x+16)\cdot(18x+42)}{(3x+7)^4}$$
$$h^\prime(x)=\frac{18x^3+84x^2+98x+72x^2+336x+392-18x^3-144x^2-288x-42x^2-336x-672}{(3x+7)^4}$$
$$h^\prime(x)=\frac{-30x^2-190x-280}{(3x+7)^4}$$
$$p(x)=\frac{1}{x}$$
$$p^\prime(x)=\frac{(1)^\prime\cdot x-1\cdot (x)^\prime}{x^2}$$
$$p^\prime(x)=\frac{0\cdot{x}-1\cdot1}{x^2}$$
$$p^\prime(x)=\frac{-1}{x^2}$$
quotient rule
$$h(x)=\frac{f(x)}{g(x)} \longrightarrow h^\prime(x)=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{(g(x))^2}$$