jpd5184 Posted January 24, 2011 Share Posted January 24, 2011 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 Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted January 24, 2011 Share Posted January 24, 2011 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. Link to comment Share on other sites More sharing options...
jpd5184 Posted January 24, 2011 Author Share Posted January 24, 2011 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) Link to comment Share on other sites More sharing options...
Fuzzwood Posted January 25, 2011 Share Posted January 25, 2011 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. Link to comment Share on other sites More sharing options...
jpd5184 Posted January 25, 2011 Author Share Posted January 25, 2011 i thought it was right to but its not the right answer. maybe it needs simplified. dont know Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted January 25, 2011 Share Posted January 25, 2011 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. 1 Link to comment Share on other sites More sharing options...
khaled Posted January 25, 2011 Share Posted January 25, 2011 (edited) 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 Link to comment Share on other sites More sharing options...
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