derivatives

[jpd5184]

the question is find the derivative of y=sqrt.(29arctan(x))

 

what i did was change it to make it y= (29arctan(x))^1/2

 

since ddx arctan = 1/1+x^2, I said the answer was (29/sqrt.(1+x^2)^1/2

[Cap'n Refsmmat]

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.

[jpd5184]

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)

[Fuzzwood]

That looks about right, you missed that the square root of the entire thing was also a function of x, for which you corrected in the 2nd post.

[jpd5184]

i thought it was right to but its not the right answer. maybe it needs simplified. dont know

[Cap'n Refsmmat]

Yes, you have to use the chain rule, because the arctan is inside a square root. It's a function inside a function; you have to use the chain rule.

[khaled]

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 January 25, 2011 by khaled