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Quotient rule with square roots

sleepless

New member
Sep 6, 2012
1
I have the answer to this problem but I am stumped as how to get there. Here it is

h(x)=e^x/5/sqrt2x^2-10x+17, i'm getting stuck moving the square root up. Help
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
I have the answer to this problem but I am stumped as how to get there. Here it is

h(x)=e^x/5/sqrt2x^2-10x+17, i'm getting stuck moving the square root up. Help
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?

\(\displaystyle \frac{\frac{e^{x}}{5}}{\sqrt{2x^2-10x+17}}\)?
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
I have the answer to this problem but I am stumped as how to get there. Here it is

h(x)=e^x/5/sqrt2x^2-10x+17, i'm getting stuck moving the square root up. Help
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)
 

soroban

Well-known member
Feb 2, 2012
409
Hello, sleepless!

I have the answer to this problem but I am stumped as how to get there.

. . h(x) = e^x/5/sqrt2x^2-10x+17.

i'm getting stuck moving the square root up. .Why do you want to do that?

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]
 

skeeter

Well-known member
MHB Math Helper
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?