Question

Derivative y=x^cos(x)

I think i am doing something completely wrong or just my steps... wondering if anyone can show step?

Answer 1

Let's see your steps. \[ln(y) = \cos(x)\cdot\ln(x)\] Now what?

Answer 2

Implicit differentiation?

Answer 3

you can do what @tkhunny suggests, but it is also important to know that the definition of \(b^x\) is \(e^{x\ln(b)}\) so you are looking at
\[e^{\cos(x)\ln(x)}\] and you take the derivative using the chair rule
you get
\[e^{\cos(x)\ln(x)}\times \frac{d}{dx}[x\cos(x)]\]

Answer 4

Implicit and Logarithmic!

Answer 5

if you say "take the log" once you find the derivative of \(\cos(x)\ln(x)\) which is where all the work is , you multiply by the original function
also i made a typo, it should be
\[e^{\cos(x)\ln(x)}\times \frac{d}{dx}[\cos(x)\ln(x)]\]

Answer 6

i think i got a mistake on your step hunny, which is what i did.. but it's probably just me

Answer 7

Note: You DO have to be careful with such functions when they meet calculators and computer math programs (Maple, Mathematica, MatthCad to name a few). You may get only positive values for 'x' unless you try hard to get the negative ones.

In full agreement with @satellite73, we shoudl also keep in mind that the transformation \[\log_{b}(a) = c \iff b^{c} = a\] for suitable a, b, and c is quite generally applicable. This makes my logarithmic form and satellites73 exponential form very much the same.

Just wade carefully thougth the chain rule, the product rule, and a little algebra.\[\ln(y) = cos(x)\cdot\ln(x)\]leads to \[\frac{y'}{y} = \frac{\cos(x)}{x} -\sin(x)\cdot\ln(x)\]