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# Quotient Rule Derivative

Top
 Sub Topics A method for finding derivative of a quotient of two functions is known to be quotient rule. The formula is applied to finding the derivative of any fraction which has a variable in both the numerator and the denominator. This formula is derived using the product rule and chain rule. Consider a function h(x) it is actually a ratio of your couple functions f(x) along with g(x).How could be the derivative of h(x) related to f(x), g(x), with their derivatives? Interior calculus, the quotient rule is really a method of choosing the derivative of your function be the quotient of a few other functions knowning that derivatives exist.

## Definition

Let f and g possibly be differentiable at x with g(x) $\neq$ 0. Then $\frac{f}{g}$ is differentiable from x and:

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

## Formula

The actual quotient rule can be a formula for choosing the derivative of a new fraction. It follows on the limit definition of derivative and is also given by:

If we've got y = $\frac{u}{v}$, where u in addition to v both are classified as the function of x and where u, v are differential operates and v $\neq$ 0. If you should calculate $\frac{dy}{ dx}$, then,

$\frac{d}{ dx}$ $\frac{u}{v}$ = $\frac{v \frac{du}{dx} - u \frac{dv}{dx}} {v^{2}}$

$\frac{dy}{ dx}$ = $\frac{d}{dx}$ ($\frac{u}{v}$)

= $\frac{vu′−uv′}{ v^{2}}$

Where by u' = $\frac{du}{dx}$ and v' = $\frac{dv}{dx}$.

## Examples

Some examples on quotient rule derivative are given below:

Example 1: Differentiate g(x) = $\frac{2^{x}}{2^{x}- 3^{x}}$
Solution:

Then g'(x) = $\frac{(2^{x} - 3^{x})D(2^{x})-(2^{x})D(2^{x}-3^{x})}{(2^{x}-3^{x})^{2}}$

= We know that D{a$^{x}$} = a$^{x}$lna

= $\frac{(2^{x}-3^{x})(2^{x}ln2)- 2^{x}(2^{x}ln2 - 3^{x}ln3)}{(2^{x}-3^{x})^{2}}$

=$\frac{2^{x}2^{x}ln2 - 3^{x}2^{x}ln2 - (2^{x}2^{x}ln2 - 2^{x}3^{x}ln3)}{(2^{x}-3^{x})^{2}}$

Recall that a$^{x}a^{x}$ = a$^{x+x}$ =a$^{2x}$

and a$^{x}b^{x} = (ab)^{x}$

= $\frac{2^{2x}ln2 - 6^{x}ln2 - (2^{2x}ln 2 - 6^{x}ln 3)}{(2^{x}-3^{x})^{2}}$

= $\frac{2^{2x}ln 2 - 6^{x}ln2 - 2^{2x}ln2 + 6^{x}ln3}{(2^{x}-3^{x})^{2}}$

=$\frac{6^{x}(ln3-ln2)}{(2^{x}-3^{x})^{2}}$

Wkt, lnA - lnB = ln($\frac{A}{B}$)

= $\frac{6^{x}ln(\frac{3}{2})}{(2^{x}-3^{x})^{2}}$

Example 2: Differentiate f(x) = $\frac{4x^{3} - 7x}{5x^{2} + 2}$

Solution:

y' = $\frac{(5x^{2}+2) D(4x^{3} - 7x) -(4x^{3} - 7x)D(5x^{2}+2)}{(5x^{2} + 2)^{2}}$

= $\frac{(5x^{2} + 2)(12x^{2} - 7) - (4x^{3}-7x)(10x)}{(5x^{2} + 2)^{2}}$

= $\frac{60x^{4} - 35x^{2} + 24x^{2} - 14 - (40x^{4} - 70x^{2})}{(5x^{2} + 2)^{2}}$

= $\frac{60x^{4} - 35x^{2} + 24x^{2} - 14 - 40x^{4} + 70x^{2}}{(5x^{2} + 2)^{2}}$

= $\frac{20x^{4} + 59x^{2} - 14}{(5x^{2} + 2)^{2}}$