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



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Posted on (Answer #1)

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")' `


`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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