Question

Find y' where y=x^(2x) *x^(x^2+2) ?

Answer 1

\[f(x) = x^{2x}\cdot x^{x^2+2} = x^{x^2+2x+2}\]

Answer 2

Factor out the top first.

Answer 3

OR NO

Answer 4

log and implicit

Answer 5

You're just trying to find y'. Then just -.... Thanks dan -.-

Answer 6

i think

Answer 7

Ther is no other way, you have a base x raised to the power of the same variable. I would think so

Answer 8

yes, I am stuck with this problem for a while. I thought of some sort of substitution, but it didn't help

Answer 9

Continuing\[
\ln(y) = (x^2+2x+2)\ln(x)
\]

Answer 10

\[
\frac{d}{dx}\ln(y) =\frac{d}{dy}\ln(y) \frac{dy}{dx}= \frac{1}{y}y'
\]

Answer 11

\[
\frac{y'}{y} = (2x+2)\ln(x) + (x^2+2x+2)\frac{1}{x}
\]

Answer 12

We can substitute \(y\) for it's function of \(x\).\[
\frac{y'}{y}=\frac{y'}{x^{x^2+2x+2}}
\]

Answer 13

This is how you do things with logarithmic differentiation.

Answer 14

Another way to do this, is to use: \[
a^x = e^{x\ln a } \implies x^{x^2+2x+2} = e^{(x^2+2x+2)\ln(x)}
\]

Answer 15

Both result in: \[
y'=\left(x^{x^2+2x+2}\right)\left((2x+2)\ln(x)+\frac{x^2+2x+2}{x}\right)
\]

Answer 16

Thank you very much again ! Now since you explained it all it is not as bad as it was..

Answer 17

I appreciate it !

Answer 18

So I didn't do anything confusing?