1
$\begingroup$

Find $y'$ for $y = x^{2\cos{x}}$

I got $y' = \left(\frac {2\cos{x}}{x} - 2\ln(x)\cos(x)\sin(x)\right)(x^{2\cos{x}})$ is that correct?

  • 0
    +1 for work shown. Though it should probably be in the question next time and not in a comment.2012-10-21

1 Answers 1

2

Nope. Your answer is a bit off. Let's do it this way:

$y=x^{2\cos{x}}=e^{2\cos{x}\ln{x}}$

hence, by the chain rule we have:

$y^{\prime}=e^{2\cos{x}\ln{x}}\left(\frac{2\cos{x}}{x}-2\sin{x}\ln{x}\right)$

and now substituting back $e^{2\cos{x}\ln{x}}=x^{2\cos{x}}$ we finally arrive at:

$y^{\prime}=x^{2\cos{x}}\left(\frac{2\cos{x}}{x}-2\sin{x}\ln{x}\right)$

EDIT: If you want to use your method:

$\ln(y)=2\cos{x}\ln{x}$

$\frac{1}{y}\cdot{y^{\prime}}=\left(\frac{2\cos{x}}{x}-2\sin{x}\ln{x}\right)$

hence:

$y^{\prime}=y\left(\frac{2\cos{x}}{x}-2\sin{x}\ln{x}\right)$

Which yields the same answer.