Finding Derivatives of Rational Functions
To find the derivative of a rational function, we will sometimes simplify the expression using algebraic techniques we have already learned.
Find the derivative of the function [latex]f\left(x\right)=\frac{3+x}{2-x}[/latex] at [latex]x=a[/latex].
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[latex]\begin{align}{f}^{\prime }\left(a\right)&=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(a+h\right)-f\left(a\right)}{h} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{\frac{3+\left(a+h\right)}{2-\left(a+h\right)}-\left(\frac{3+a}{2-a}\right)}{h}&& \text{Substitute }f\left(a+h\right)\text{ and }f\left(a\right) \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{\left(2-\left(a+h\right)\right)\left(2-a\right)\left[\frac{3+\left(a+h\right)}{2-\left(a+h\right)}-\left(\frac{3+a}{2-a}\right)\right]}{\left(2-\left(a+h\right)\right)\left(2-a\right)\left(h\right)}&& \text{Multiply numerator and denominator by }\left(2-\left(a+h\right)\right)\left(2-a\right) \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{\bcancel{\left(2-\left(a+h\right)\right)}\left(2-a\right)\left(\frac{3+\left(a+h\right)}{\bcancel{\left(2-\left(a+h\right)\right)}}\right)-\left(2-\left(a+h\right)\right)\bcancel{\left(2-a\right)}\left(\frac{3+a}{\bcancel{2-a}}\right)}{\left(2-\left(a+h\right)\right)\left(2-a\right)\left(h\right)}&& \text{Distribute and cancel} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{6 - 3a+2a-{a}^{2}+2h-ah - 6+3a+3h - 2a+{a}^{2}+ah}{\left(2-\left(a+h\right)\right)\left(2-a\right)\left(h\right)}&& \text{Multiply} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{5\cancel{h}}{\left(2-\left(a+h\right)\right)\left(2-a\right)\left(\cancel{h}\right)}&& \text{Combine like terms} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{5}{\left(2-\left(a+h\right)\right)\left(2-a\right)}&& \text{Cancel like factors} \\ &=\frac{5}{\left(2-\left(a+0\right)\right)\left(2-a\right)}=\frac{5}{\left(2-a\right)\left(2-a\right)}=\frac{5}{{\left(2-a\right)}^{2}}&& \text{Evaluate the limit} \end{align}[/latex]
Find the derivative of the function [latex]f\left(x\right)=\frac{10x+11}{5x+4}[/latex] at [latex]x=a[/latex].
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[latex]\begin{align}{f}^{\prime }\left(a\right)=\frac{-15}{{\left(5a+4\right)}^{2}}\end{align}[/latex]
Finding Derivatives of Functions with Roots
To find derivatives of functions with roots, we use the methods we have learned to find limits of functions with roots, including multiplying by a conjugate.
Find the derivative of the function [latex]f\left(x\right)=4\sqrt{x}[/latex] at [latex]x=36[/latex].
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We have
[latex]\begin{align}{f}^{\prime }\left(a\right)&=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(a+h\right)-f\left(a\right)}{h} \\ &=\underset{h\to 0}{\mathrm{lim}}\frac{4\sqrt{a+h}-4\sqrt{a}}{h}&& \text{Substitute }f\left(a+h\right)\text{ and }f\left(a\right) \end{align}[/latex]
Multiply the numerator and denominator by the conjugate: [latex]\frac{4\sqrt{a+h}+4\sqrt{a}}{4\sqrt{a+h}+4\sqrt{a}}[/latex].
[latex]\begin{align}{f}^{\prime }\left(a\right)&=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4\sqrt{a+h}-4\sqrt{a}}{h}\right)\cdot \left(\frac{4\sqrt{a+h}+4\sqrt{a}}{4\sqrt{a+h}+4\sqrt{a}}\right) \\ &=\underset{h\to 0}{\mathrm{lim}}\left(\frac{16\left(a+h\right)-16a}{h\left(4\sqrt{a+h}+4\sqrt{a}\right)}\right)&& \text{Multiply}. \\ &=\underset{h\to 0}{\mathrm{lim}}\left(\frac{16a+16h-16a}{h\left(4\sqrt{a+h}+4\sqrt{a}\right)}\right) && \text{Distribute and combine like terms}. \\ &=\underset{h\to 0}{\mathrm{lim}}\left(\frac{16\cancel{h}}{\cancel{h}\left(4\sqrt{a+h}+4\sqrt{a}\right)}\right)&& \text{Simplify}. \\ &=\underset{h\to 0}{\mathrm{lim}}\left(\frac{16}{4\sqrt{a+h}+4\sqrt{a}}\right)&& \text{Evaluate the limit by letting }h=0. \\ &=\frac{16}{8\sqrt{a}}=\frac{2}{\sqrt{a}} \\ {f}^{\prime }\left(36\right)&=\frac{2}{\sqrt{36}} =\frac{2}{6} =\frac{1}{3} && \text{Evaluate the derivative at }x=36. \end{align}[/latex]
Find the derivative of the function [latex]f\left(x\right)=9\sqrt{x}[/latex] at [latex]x=9[/latex].
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[latex]\frac{3}{2}[/latex]