Derivative of a Function: y= (2x)^(1/2) + (2/x)^(1/2), etc

[coolaid2u]

I'm having some problems finding the dervative of the funtions below.. I end up with different answers each time I do them .

1) y= (x^2-5)^(1/2) * (x^2+3)^(1/3)

2) y= (2x)^(1/2) + (2/x)^(1/2)

3) y= (4x^2+7)^2 * (3x^3+1)^4

I'd be happy if anyone could help! : )

 

[stapel]

I end up with different answers each time I do them

We'll be glad to help you find any errors, but you'll need to show your work first.

Thank you.

Eliz.

 

[soroban]

Re: Derivative of a Function!

Hello, coolaid2u!

You must apply the Product Rule and Chain Rule very carefully . . .

\(\displaystyle \L y\:=\x^2\,-\,5)^{\frac{1}{2}}\cdot(x^2\,+\,3)^{\frac{1}{3}}\)

We have: \(\displaystyle \:\L y'\;=\;(x^2\,-\,5)^{\frac{1}{2}}\cdot\frac{1}{3}(x^2\,+\,3)^{-\frac{2}{3}}\cdot(2x)\:+\x^2\,+\,3)^{\frac{1}{3}}\cdot\frac{1}{2}(x^2\,-\,5)^{-\frac{1}{2}}\cdot(2x)\)

. . . \(\displaystyle \L y' \;=\;\frac{2}{3}x(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{-\frac{2}{3}}\:+\:x(x^2\,-\,5)^{-\frac{1}{2}}(x^2\,+\,3)^{\frac{1}{3}}\)

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Most textbooks simplify these answers beyond all recognition.
Here's how it's done . . .

Multiply "top and bottom" by \(\displaystyle 3(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{\frac{2}{3}}\)

\(\displaystyle \L\frac{3(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{\frac{2}{3}} }{3(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{\frac{2}{3}}}\,\cdot\,\frac{\frac{2}{3}x(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{-\frac{2}{3}} \,+\,x(x^2\,-\,5)^{-\frac{1}{2}}(x^2\,+\,3)^{\frac{1}{3}}}{1}\)

. . \(\displaystyle \L= \;\frac{2x(x^2\,-\,5)\,+\,3x(x^2\,+\,3)}{3(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,e)^{\frac{2}{3}}} \;=\;\frac{2x^3\,-\,10x\,+\,3x^2\,+\,9x}{3(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{\frac{2}{3}}}\)

. . \(\displaystyle \L=\;\frac{5x^3\,-\,x}{3(x^2\,-\,5)^{\frac{1}{2}}(x^2\,+\,3)^{\frac{2}{3}}}\)

\(\displaystyle \L y \:=\: (2x)^{\frac{1}{2}}\,+\,\left(\frac{2}{x}\right)^{\frac{1}{2}}\)

I would simplify the function first . . .

We have: \(\displaystyle \L\;y \;= \;(2x)^{\frac{1}{2}}\,+\,\left(\frac{2}{x}\right)^{\frac{1}{2}} \;=\;\sqrt{2}\cdot x^{\frac{1}{2}}\,+\,\sqrt{2}\cdot x^{-\frac{1}{2}}\)

Go for it!