Quotient Rule - Top Future Point

In calculus , the quotient rule is a method of finding the derivative of a function that is the ratio of two differential functions. Let g and h be both differentiable and the quotient rule states that f ( x ) is a derivative of

f(x)=g(x)/h(x),}{\displaystyle h(x)\neq 0.

f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.

Example

A basic example:

{\begin{aligned}{\frac {d}{dx}}{\frac {e^{x}}{x^{2}}}&={\frac {\left({\frac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\frac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\frac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}.\end{aligned}}

The quotient rule can be used to find the derivative of

f(x)=\tan x={\tfrac {\sin x}{\cos x}}

{\begin{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}{\frac {\sin x}{\cos x}}\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}

Evidence

Derived definition and proof from boundary properties

The definition of the LET derivative and applying the properties of the limits provides the following proof.

f(x)={\frac {g(x)}{h(x)}}.

{\begin{aligned}f'(x)&=\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {g(x+k)}{h(x+k)}}-{\frac {g(x)}{h(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k\cdot h(x)h(x+k)}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{h(x)h(x+k)}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)+g(x)h(x)-g(x)h(x+k)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)}{k}}-\lim _{k\to 0}{\frac {g(x)h(x+k)-g(x)h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(h(x)\lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}-g(x)\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}

Proof using implicit differentiation

Lets then solve for for the product rule then gives and substituting back for lets:

{\displaystyle f(x)={\frac {g(x)}{h(x)}},}{\displaystyle g(x)=f(x)h(x).}{\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x).}f'(x)f(x)

{\begin{aligned}f'(x)&={\frac {g'(x)-f(x)h'(x)}{h(x)}}\\&={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}

Proof using chain rule

Let ‘s then rule out the product

f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.

f'(x)=g'(x)h(x)^{-1}+g(x)\cdot {\frac {d}{dx}}(h(x)^{-1}).

To evaluate the derivative in the second term, apply the power rule in conjunction with the chain rule :

f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).

Finally, rewrite as fractions and combine the terms to get

{\begin{aligned}f'(x)&={\frac {g'(x)}{h(x)}}-{\frac {g(x)h'(x)}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}

Higher order formula

Implicit differentiation can be used to calculate the nth derivative of the quotient ( partly in terms of its first n -1 derivative). For example, differentiating twice (resulting in ) and then solving yields

{\displaystyle fh=g}{\displaystyle f''h+2f'h'+fh''=g''}f''

f''=\left({\frac {g}{h}}\right)''={\frac {g''-2f'h'-fh''}{h}}.

What is the rule of division?

When a number or digit is subtracted more than once in a number or digit, it is called division. The number of times a number or digit is divided, the same number of times has to be divided.

What is the dividend divisor quotient remainder?

Division
Division is that operation in mathematics by which the product of two numbers and given one of these numbers, the other is found. The given product is called the ‘dividend’ (dividend or numerator), the given number is called ‘divisor or denominator’ and the required number is called ‘quotient’.