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Differentiate the following function a) g(x)=\frac{x^2+5}{x^2-4} b) h(x) =\frac{x^2+4x+3}{\sqrt{x}}

Question:

Differentiate the following function

a) {eq}g(x)=\frac{x^2+5}{x^2-4} {/eq}

b) {eq}h(x) =\frac{x^2+4x+3}{\sqrt{x}} {/eq}

Quotient Rule:

The quotient rule of derivatives is applied when we are given with a function containing 2 or more sub functions in fractions. Let the 2 sub functions are given as f(x) and g(x) and their fraction is given as a function h(x) such that:

{eq}\displaystyle { h(x) = \frac{f(x)}{g(x)} } {/eq}

Then the quotient rule can be applied as:

{eq}\displaystyle { h'(x) = \frac{(g(x))(f'(x)) - (g'(x))(f(x)) }{g(x)} } {/eq}

Some important formulas of differentiation are:

{eq}\displaystyle { \frac{d}{dx} \sqrt x = \frac{1}{2 \sqrt x} \\ \frac{d}{dx} x^n = nx^{n-1} \\ \frac{d}{dx} \ln (x) = \frac{1}{x} } {/eq}

Answer and Explanation:

(a)

{eq}\displaystyle { g(x)=\frac{x^2+5}{x^2-4} } {/eq}

Differentiating the given function:

{eq}\displaystyle { g'(x)=\frac{(x^2-4)(x^2+5)'-(x^2-4)'(x^2+5)}{(x^2-4)^2} \ \ \ \ \ \ \text{(Quotient rule)} \\ = \frac{(x^2-4)(2x)-(2x)(x^2+5)}{(x^2-4)^2} \ \ \ \ \ \ \text{(Power rule)} \\ = \frac{ (2x) [ (x^2-4)-(x^2+5) ] }{(x^2-4)^2} \\ = \frac{ (2x) [ (x^2-4-x^2-5) ] }{(x^2-4)^2} \\ = \frac{ -18x }{(x^2-4)^2} } {/eq}

which is the derivative of the given function.


(b)

{eq}\displaystyle { h(x) =\frac{x^2+4x+3}{\sqrt{x}} } {/eq}

Differentiating the given function:

{eq}\displaystyle { h'(x)=\frac{(\sqrt{x})(x^2+4x+3)'-(\sqrt{x})'(x^2+4x+3)}{(\sqrt{x})^2} \ \ \ \ \ \ \text{(Quotient rule)} \\ = \frac{(\sqrt{x})(2x+4)-(\frac{1}{2\sqrt{x}})(x^2+4x+3)}{x} \ \ \ \ \ \ \text{(Power rule)} \\ = \frac{2x+4}{\sqrt x}-\frac{1}{2x\sqrt{x}}(x^2+4x+3) } {/eq}

which is the derivative of the given function.


Learn more about this topic:

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When to Use the Quotient Rule for Differentiation

from Math 104: Calculus

Chapter 8 / Lesson 8
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