How do you differentiate $x^{\cos(x)}$

Question

How do you differentiate $x^{\cos(x)}$

I encounter this problem in my homework but I don't know how to differentiate it. Do you need use logarithmic differentiation to do this?

Thank you!

Answer 1

Let $$y=x^{\cos(x)}$$ then we get by taking the logarithm $$\ln(y)=\cos(x)\ln(x)$$ and now by the chain rule we obtain: $$\frac{y'(x)}{y(x)}=-\sin(x)\ln(x)+\frac{\cos(x)}{x}$$ Can you finish?

Answer 2

use $x^{\cos(x)} = e^{\ln(x^{\cos(x)})}= e^{\cos(x)\ln(x)}$.

Answer 3

Hint: $x^{\cos(x)}=e^{\cos(x)\ln(x)}$.

Answer 4

Let $y=x^{\cos(x)}$. Now take the log of both sides:

$$\log(y) = \cos(x) \log(x). $$

Now differentiate both sides (with the product rule):

$$ \frac{y'}{y} = -\sin(x) \log(x) + \cos(x) \frac{1}{x}.$$

Note that $y'$ (the expression we seek) equals $$ y' = y\left[ -\sin(x) \log(x) + \cos(x) \frac{1}{x} \right]. $$

Replace the original expression for $y$ into the above and you're done.

Answer 5

Usually functions with a variable in the base and exponent are best done by passing to the logarithm: $$x^{\cos x}= e^{\cos x \log x}$$ Which by the chain rule has derivative $$(\cos x \log x)’ e^{\cos x \log x}= (\frac{\cos x}{x} -\sin x \log x) e^{\cos x \log x} $$

Answer 6

The function is defined on $(0,\infty)$.

Start with: $$x^{\cos x}=e^{\ln x^{\cos x}}=e^{{\cos x}\ln x}$$

Now apply chain rule:$$\cdots=e^{{\cos x}\ln x}\left[-\sin x\ln x+\frac{\cos x}{x}\right]=x^{\cos x}\left[-\sin x\ln x+\frac{\cos x}{x}\right]$$