Find the first derivative of y=(x+x^-1)^2

Question

perl · Answer

did you try chain rule

perl · Answer

y = u^n

dy/dx = n * u^(n-1) * du/dx

anonymous · Answer

Yeah i tried it but i keep getting an answer that is not on the matching answer sheet so i think im doing something wrong.

perl · Answer

dy/dx = 2 * ( x + x^-1 )^1 * ( 1 - x^-2 )

jhannybean · Answer

$$f(x) = (x+x^{-1})^2 = \left(x+\frac{1}{x}\right)^2 = \left(\frac{x^2+1}{x}\right)^2$$

jhannybean · Answer

That might make things easier, maybe.

anonymous · Answer

Here is how I like to look at it: $$
\begin{split}
dy 
&= d[(x+x^{-1})^2] &\text{let}\;u=(x+x^{-1}) \
&=d(u^2)\
&= 2udu\ 
&= 2(x+x^{-1})d(x+x^{-1})
\end{split}

$$

anonymous · Answer

It has to be simplified farther than that and the chain rule has to be used but thank you! I keep getting confused when i try to simplify it past what Perl got.

jhannybean · Answer

$$f'(x) = 2\left(\frac{x^2+1}{x}\right) \cdot \left[\frac{2x(x)-1(x^2+1)}{x^2}\right]$$$$f'(x) =2\left(\frac{x^2+1}{x}\right)\cdot\left[\frac{2x^2-x^2-1}{x^2}\right]$$

jhannybean · Answer

That way work.

jhannybean · Answer

Then just simplify what is in the brackets (my explanation)

Albeit, there are various methods of solving something as trivial as this question.

anonymous · Answer

Here is my complete version: $$
\begin{split}
dy 
&= d[(x+x^{-1})^2] &\text{let}\;u=(x+x^{-1}) \
&=d(u^2)\
&= 2udu\ 
&= 2(x+x^{-1})d(x+x^{-1})\
&=2(x+x^{-1})(1-x^{-2})dx
\end{split}\
\;
\
\implies
\frac{dy}{dx} = 2(x+x^{-1})(1-x^{-2})

$$

jhannybean · Answer

If you simplify @wio 's answer in making all variables positive, you will end up with my answer, after you have simplified it. :-) I love derivatives.

jhannybean · Answer

http://www.wolframalpha.com/input/?i=d%2Fdx+f%28x%29%3D%28x%2Bx%5E-1%29%5E2