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find the derivatives  y= [ln (x^2+3)]^3/2

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good2beme | Student, Undergraduate | (Level 1) Honors

Posted June 15, 2012 at 2:08 AM via web

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find the derivatives 

y= [ln (x^2+3)]^3/2

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

Posted June 15, 2012 at 4:43 AM (Answer #1)

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This is a chain rule problem.

The outermost function is `("stuff")^((3)/(2))`
The next most "outer" function is `"ln"("stuff")`
The innermost function is `x^2+3`

The chain rule says:

`d/dx f(g(x))=f'(g(x))g'(x)`

In simple terms, the chain rule says:


` d/dx ("stuff")^((3)/(2)) = (3)/(2)("stuff")^((1)/(2))("stuff")' `
`d/dx "ln"("stuff") = (1)/("stuff")*("stuff")' `


So:

`d/dx (x^2+3) = 2x `
`d/dx "ln"(x^2+3) = (1)/(x^2+3)*2x`
`d/dx ("ln" (x^2+3))^((3)/(2))= (3)/(2) ("ln" (x^2+3))^((1)/(2))*(1)/(x^2+3)*2x`

`=3 sqrt("ln" (x^2+3)) * (x)/(x^2+3)`

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