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Quotient rule with square roots
- Thread starter sleepless
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- Jan 26, 2012
- 4,041
Hi sleepless,
Welcome to MHB! I'm not exactly sure what problem you are trying to solve. Try to be very precise with parentheses. Is this what you want to take the derivative of?
\(\displaystyle \frac{\frac{e^{x}}{5}}{\sqrt{2x^2-10x+17}}\)?
- Mar 1, 2012
- 249
Or is it
$ h(x) = \dfrac{e^{x/5}}{\sqrt{2x^2-10x+17}}$
With square roots note that $\sqrt{x} = x^{1/2}$ and don't forget the chain rule where appropriate (which you will need in this example)
Hello, sleepless!
I'll take a guess as to what the problem is . . .
. . [tex]h(x) \;=\;\frac{e^{\frac{x}{5}}}{\sqrt{2x^2 - 10x + 17}} \;=\;\frac{e^{\frac{1}{5}x}}{(2x^2 - 10x + 17)^{\frac{1}{2}}}[/tex]
[tex]\text{Quotient Rule:}[/tex]
. . [tex]h'(x) \;=\; \frac{(2x^2-10x+17)^{\frac{1}{2}}\cdot e^{\frac{1}{5}x} \!\cdot\!\frac{1}{5} \;-\; e^{\frac{1}{5}x}\!\cdot\!\frac{1}{2}(2x^2-10x+17)^{-\frac{1}{2}}(4x-10)} {2x^2 - 10x + 7}[/tex]
. . . . . . [tex]=\;\frac{\frac{1}{5}e^{\frac{x}{5}}(2x^2 - 10x + 17)^{\frac{1}{2}} \;-\; e^{\frac{x}{5}}(2x-5)(2x^2-10x+17)^{-\frac{1}{2}}}{2x^2-10x+17} [/tex]
Multiply numerator and denominator by [tex](2x^2-10x + 17)^{\frac{1}{2}}[/tex]
. . [tex]h'(x) \;=\;\frac{\frac{1}{5}e^{\frac{x}{5}}(2x^2-10x + 17) \;-\; e^{\frac{x}{5}}(2x-5)}{(2x^2-10x+17)^{\frac{3}{2}}} [/tex]
. . . . . . [tex]=\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot\!\frac{(2x^2 - 10x + 17) \;-\; 5(2x-5)}{(2x^2-10x+17)^{\frac{3}{2}}}[/tex]
. . . . . . [tex]=\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot\!\frac{2x^2 - 10x + 17 - 10x + 25}{(2x^2-10x+17)^{\frac{3}{2}}} [/tex]
. . . . . . [tex]=\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot \!\frac{2x^2-20x + 42}{(2x^2-20x+17)^{\frac{3}{2}}}[/tex]
. . . . . . [tex]=\;\tfrac{2}{5}e^{\frac{x}{5}}\!\cdot\!\frac{x^2-10x + 21}{(2x^2-10x+17)^{\frac{3}{2}}}[/tex]
I'll take a guess as to what the problem is . . .
. . [tex]h(x) \;=\;\frac{e^{\frac{x}{5}}}{\sqrt{2x^2 - 10x + 17}} \;=\;\frac{e^{\frac{1}{5}x}}{(2x^2 - 10x + 17)^{\frac{1}{2}}}[/tex]
[tex]\text{Quotient Rule:}[/tex]
. . [tex]h'(x) \;=\; \frac{(2x^2-10x+17)^{\frac{1}{2}}\cdot e^{\frac{1}{5}x} \!\cdot\!\frac{1}{5} \;-\; e^{\frac{1}{5}x}\!\cdot\!\frac{1}{2}(2x^2-10x+17)^{-\frac{1}{2}}(4x-10)} {2x^2 - 10x + 7}[/tex]
. . . . . . [tex]=\;\frac{\frac{1}{5}e^{\frac{x}{5}}(2x^2 - 10x + 17)^{\frac{1}{2}} \;-\; e^{\frac{x}{5}}(2x-5)(2x^2-10x+17)^{-\frac{1}{2}}}{2x^2-10x+17} [/tex]
Multiply numerator and denominator by [tex](2x^2-10x + 17)^{\frac{1}{2}}[/tex]
. . [tex]h'(x) \;=\;\frac{\frac{1}{5}e^{\frac{x}{5}}(2x^2-10x + 17) \;-\; e^{\frac{x}{5}}(2x-5)}{(2x^2-10x+17)^{\frac{3}{2}}} [/tex]
. . . . . . [tex]=\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot\!\frac{(2x^2 - 10x + 17) \;-\; 5(2x-5)}{(2x^2-10x+17)^{\frac{3}{2}}}[/tex]
. . . . . . [tex]=\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot\!\frac{2x^2 - 10x + 17 - 10x + 25}{(2x^2-10x+17)^{\frac{3}{2}}} [/tex]
. . . . . . [tex]=\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot \!\frac{2x^2-20x + 42}{(2x^2-20x+17)^{\frac{3}{2}}}[/tex]
. . . . . . [tex]=\;\tfrac{2}{5}e^{\frac{x}{5}}\!\cdot\!\frac{x^2-10x + 21}{(2x^2-10x+17)^{\frac{3}{2}}}[/tex]
- Mar 1, 2012
- 696
[tex]y = \frac{e^{x/5}}{\sqrt{2x^2-10x+17}}[/tex]
[tex]\ln{y} = \frac{x}{5} - \frac{1}{2}\ln(2x^2-10x+17)[/tex]
[tex]\frac{y'}{y} = \frac{1}{5} - \frac{2x-5}{2x^2-10x+17}[/tex]
[tex]\frac{y'}{y} = \frac{2x^2-20x+42}{5(2x^2-10x+17)}[/tex]
[tex]y' = \frac{e^{x/5}}{\sqrt{2x^2-10x+17}} \cdot \frac{2x^2-20x+42}{5(2x^2-10x+17)}[/tex]
[tex]y' = \frac{2e^{x/5}(x^2-10x+21)}{5(2x^2-10x+17)^{\frac{3}{2}}}[/tex]
don't you love logarithmic differentiation?
[tex]\ln{y} = \frac{x}{5} - \frac{1}{2}\ln(2x^2-10x+17)[/tex]
[tex]\frac{y'}{y} = \frac{1}{5} - \frac{2x-5}{2x^2-10x+17}[/tex]
[tex]\frac{y'}{y} = \frac{2x^2-20x+42}{5(2x^2-10x+17)}[/tex]
[tex]y' = \frac{e^{x/5}}{\sqrt{2x^2-10x+17}} \cdot \frac{2x^2-20x+42}{5(2x^2-10x+17)}[/tex]
[tex]y' = \frac{2e^{x/5}(x^2-10x+21)}{5(2x^2-10x+17)^{\frac{3}{2}}}[/tex]
don't you love logarithmic differentiation?