Subject: Calculus

# Quotient Rule

One very important theorem on derivative is the Quotient Rule which is presented below. This rule best applies to functions that are expressed as a quotient.

## Theorem: (Derivative of a Quotient)

If h and g are differentiable at x such that f(x)=\frac{g(x)}{h(x)}, where h(x)\neq 0, then the derivative of f at x is given by

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

### Illustration

Differentiate f(x)=\frac{x^{2}}{2x}. Solution: Again, we know both the derivative of the denominator and numerator of f(x) but according to the theorem above, we can't just take the directly the quotient of the individual derivatives as the derivative of f. Instead,

 f'(x) = \frac{2xD_{x}(x^{2}) - x^{2}D_{x}(2x)}{[2x]^{2}} = \frac{2x \cdot 2x - x^{2} \cdot 2}{(2x)^{2}} = \frac{4x^{2}-2x^{2}}{[2x]^{2}} = \frac{2x^{2}}{4x^{2}} = \frac{1}{2}

To help us remember the formula in differentiating a quotient function, remember the following mnemonics: f'(x) is low D-high minus high D-low over low squared. "High" means the numerator term while "low" represents the term in the denominator. "D-high" means the derivative of the numerator and "D-low" means differentiating the denominator. So that's it. You can now easily recall the formula of the Quotient rule.

### Example #1

Find the derivative of f when f(x)=\frac{x+1}{x+2}.

Proceeding directly, we get

f'(x)=\frac{(x+2)D_x [x+1] -(x+1)D_x [x+2] }{(x+2)^2}
= \frac{x+2 - x -1}{(x+2)^2}
= \frac{1}{(x+2)^2}

Thus,

f'(x)=\frac{1}{(x+2)^2}

### Example #2

Find the derivative of the function f(x)=\frac{x}{\sin{x}}.

Using quotient rule, we have

f'(x)=\frac{\sin{x}D_x [x] - x D_x [\sin{x}]}{\sin^2{x}}
=\frac{\sin{x}- x\cos{x}}{\sin^2{x}}

Thus,

=\frac{\sin{x}- x\cos{x}}{\sin^2{x}}

### Example #3

Using Product Rule find the derivative of f when f(x)=\frac{ \sin{x}}{\cos{x}}.

Using the same techniques in the previous examples, we have

f'(x) = \frac{\cos{x}\cdot D_x[\sin{x}]-\sin{x}D_x[\cos{x}]}{\cos^2{x}}
= \frac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}
=\frac{1}{\cos^2{x}}
=\sec^2{x}

Thus,

f'(x)=\sec^2{x}

Notice that f(x)=\frac{ \sin{x}}{\cos{x}} can also be written as f(x)=\tan{x}. By looking at the table of derivatives of trigonometric functions, we can directly get the derivative of tangent which is \sec^2{x}. This just verifies our result using quotient rule.

### Example #4

Find the derivative of the function f when f(x)=\frac{4x}{e^{2x}}.

By virtue of the quotient rule, we proceed with direct differentiation.

f'(x)=\frac{e^{2x}\cdot D_x[4x] - 4x\cdot D_x[e^{2x}]}{e^{4x}}
=\frac{e^{2x}\cdot 4 - 8x\cdot e^{2x}}{e^{4x}}
=\frac{e^{2x}[ 4 - 8x]}{e^{4x}}
=\frac{ 4 - 8x}{e^{2x}}

Thus,

f'(x)=\frac{ 4 - 8x}{e^{2x}}

### Example #5

Solve for the derivative of f(x)=\frac{x}{y^2}.

This is just direct differentiation, no more quotient rule involved.

f'(x)=\frac{1}{y^2}\cdot D_x [x]
=\frac{1}{y^2}\cdot (1)

Thus,

f'(x)=\frac{1}{y^2}

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