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?
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?
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.