What is the derivative of #x^(cosx)#?

Question

Ezekiel Alexander · Accepted Answer

#(dy)/dx = x^{cos(x)} (cos(x)/x - sin(x) log_e x)# #y = x^{cos(x)} equiv log_e y = cos(x) log_e x# Deriving the log transformed equation we have #dy/y = -sin(x)log_e x dx+cos(x)/x dx# grouping #(dy)/dx = y(cos(x)/x - sin(x) log_e x)# and finally #(dy)/dx = x^{cos(x)} (cos(x)/x - sin(x) log_e x)#

Aurora Alvarado · Answer

undefined To find the derivative of $x^{\cos(x)}$, we can use the chain rule. Let $f(x) = x^{\cos(x)}$. Then, $f'(x) = \frac{d}{dx} \left(x^{\cos(x)}\right)$.

Using the chain rule, we have:

$$f'(x) = \frac{d}{dx} \left(x^{\cos(x)}\right) = \frac{d}{dx} \left(e^{\ln(x^{\cos(x)})}\right)$$

$$= \frac{d}{dx} \left(e^{\cos(x) \ln(x)}\right)$$

$$= e^{\cos(x) \ln(x)} \left(\frac{d}{dx} (\cos(x) \ln(x))\right)$$

$$= e^{\cos(x) \ln(x)} \left(\frac{d}{dx} \cos(x) \cdot \ln(x) + \cos(x) \cdot \frac{d}{dx} \ln(x)\right)$$

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

So, the derivative of $x^{\cos(x)}$ with respect to $x$ is $e^{\cos(x) \ln(x)} \left(-\sin(x) \ln(x) + \frac{\cos(x)}{x}\right)$.