06 3quotientrule

# The Quotient Rule

Quotient means "the result of dividing" so use this when two functions have been divided (or are in a fraction).

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… … If … … $y(x)=\dfrac{u(x)}{v(x)}$
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… … Then … $\dfrac{dy}{dx}=\dfrac{v\cdot{u'}-u\cdot{v}'}{v^2}$

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### Example 1

Find the derivative of each of the following with respect to x:

a)$y=\dfrac{\sqrt{x}}{1+x}$
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b)$y=\dfrac{\log_e(x)}{\sin(x)}$
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Solution:

a)$y=\dfrac{\sqrt{x}}{1+x}$

… … … Let $u=x^{\frac{1}{2}} \qquad \text{ and } \quad \;v=1+x$

… … … … $u'=\dfrac{1}{2}x^{-\frac{1}{2}}\quad \text{ and }\quad v'=1$
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… … … $\dfrac{dy}{dx}=\dfrac{v\cdot{u'}-u\cdot{v}'}{v^2}$
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… … … … $=\dfrac{(1+x)\cdot{\frac{1}{2}x^{-\frac{1}{2}}}-x^{\frac{1}{2}}\cdot{1}}{(1+x)^2}$

… … … … $=\dfrac{{\frac{1}{2}x^{-\frac{1}{2}}(1+x)}-x^{\frac{1}{2}}} {(1+x)^2}\times\dfrac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}$

… … … … $=\dfrac{(1+x)-2x}{2x^{\frac{1}{2}}(1+x)^2}$

… … … … $=\dfrac{1-x}{2 \sqrt{x}(1+x)^2}$

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b)$y=\dfrac{\log_e(x)}{\sin(x)}$

… … … Let $u=\log_e(x)\ \quad \text{ and }\quad v=\sin(x)$

… … … … . $u'=\dfrac{1}{x} \qquad \text{ and } \quad v'=\cos(x)$

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… … … $\dfrac{dy}{dx}=\dfrac{v\cdot{u'}-u\cdot{v}'}{v^2}$
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… … … … $=\dfrac{\sin(x)\cdot{\dfrac{1}{x}}-\log_e(x)\cdot{\cos(x)}}{ \big[ \sin(x) \big]^2}$

… … … … $=\dfrac{\sin(x)\cdot{\dfrac{1}{x}} - \log_e(x)\cdot{\cos(x)}}{ \big[ \sin(x) \big]^2 }\times{\dfrac{x}{x}}$

… … … … $=\dfrac{\sin(x)-x \log_e(x)\cdot{\cos(x)}}{x \sin^2(x)}$

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