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derivatives


jpd5184

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Let me get this straight. You have:

 

[math]y = \sqrt{29 \tan^{-1}(x)}[/math]

 

So you made:

 

[math]y = (29 \tan^{-1}(x))^{\frac{1}{2}}[/math]

 

and said the answer is:

 

[math]\frac{dy}{dx} = \left( \frac{29}{\sqrt{1+x^2}} \right)^{\frac{1}{2}}[/math]

 

Right? I'm not sure I'm understanding your final equation right.

 

In any case, check how you're applying the chain rule. Your answer should still have an [imath]\tan^{-1}[/imath] in it.

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you are right with everything you said.

 

do i have to use the chain rule. i thought that because ddx arctan(x) = 1/x^2 + 1 that i could just substitute that in.

 

 

so if i use chain rule then it would be (1/2(29arctan(x))^-1/2) (29/x^2 + 1)

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Shouldn't the derivative of f(g(x)) = f'(g(x)) * g'(x) ..?

 

[imath]f(x) = \sqrt{x}[/imath]

 

[imath]g(x) = 29 \tan^{-1}{(x)}[/imath]

 

[imath]\frac{dy}{dx} f(g(x)) = f^{'}(g(x)) \times g^{'}(x)[/imath]

 

[imath]\frac{dy}{dx} \sqrt{29 \tan^{-1}{(x)}} = {(\sqrt{29 \tan^{-1}{(x)}})}^{'} \times {(29 \tan^{-1}{(x)})}^{'}[/imath]

 

[latex]

= \frac{1}{2 (\sqrt{29 \tan^{-1}{(x)}}) } \times \frac{29}{1 + x^2}

[/latex]

 

[latex]

= \frac{29}{2 (\sqrt{29 \tan^{-1}{(x)}}) \times (1 + x^2) }

[/latex]

Edited by khaled
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