MHB How Do You Apply the Quotient Rule with Square Roots?

[sleepless]

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]

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?

[math]\frac{\frac{e^{x}}{5}}{\sqrt{2x^2-10x+17}}[/math]?

 

[SuperSonic4]

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]

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 . . .

. . 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}}}\text{Quotient Rule:}

. . 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}

. . . . . . =\;\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}

Multiply numerator and denominator by (2x^2-10x + 17)^{\frac{1}{2}}

. . 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}}}

. . . . . . =\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot\!\frac{(2x^2 - 10x + 17) \;-\; 5(2x-5)}{(2x^2-10x+17)^{\frac{3}{2}}}

. . . . . . =\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot\!\frac{2x^2 - 10x + 17 - 10x + 25}{(2x^2-10x+17)^{\frac{3}{2}}}

. . . . . . =\;\tfrac{1}{5}e^{\frac{x}{5}}\!\cdot \!\frac{2x^2-20x + 42}{(2x^2-20x+17)^{\frac{3}{2}}}

. . . . . . =\;\tfrac{2}{5}e^{\frac{x}{5}}\!\cdot\!\frac{x^2-10x + 21}{(2x^2-10x+17)^{\frac{3}{2}}}

 

[skeeter]

y = \frac{e^{x/5}}{\sqrt{2x^2-10x+17}}

\ln{y} = \frac{x}{5} - \frac{1}{2}\ln(2x^2-10x+17)

\frac{y'}{y} = \frac{1}{5} - \frac{2x-5}{2x^2-10x+17}

\frac{y'}{y} = \frac{2x^2-20x+42}{5(2x^2-10x+17)}

y' = \frac{e^{x/5}}{\sqrt{2x^2-10x+17}} \cdot \frac{2x^2-20x+42}{5(2x^2-10x+17)}

y' = \frac{2e^{x/5}(x^2-10x+21)}{5(2x^2-10x+17)^{\frac{3}{2}}}

don't you love logarithmic differentiation?