Question

differentiate y=x^(e^x)

Answer 1

I'm first going to assume that you mean the following:\[y=x ^{e^x}\]

Answer 2

If this is the case, we'd be much better off using a technique called "logarithmic differentiation." Take the natural log (ln ) of both sides of \[y=x ^{e ^{x}}\]... what does that give you? Then differentiate the resulting equation with respect to x.

Answer 3

Can I like let u =e^x

Answer 4

probably not. because you have x as the base

one "trick" is take the ln of both sides
\[ \ln(y) = e^x \ln(x) \]
and take the derivative with respect to x

Answer 5

on the left side you get
\[ \frac{d}{dx} \ln(y) = \frac{1}{y} \frac{dy}{dx} \]

Answer 6

or, using \[ y=x ^{e ^{x}} \]
\[ \frac{d}{dx} \ln(y) = \frac{1}{y} \frac{dy}{dx} = x ^{-e ^{x}}\frac{dy}{dx} \]

Answer 7

I got it, Thank you

Answer 8

Hello Mr Phi...I have question that how u got x^(-e^x)

Answer 9

I only differentiated the left side.
starting from the beginning
\[ y=x^{e^x} \\ \ln(y) = e^x \cdot \ln(x) \\ \frac{d}{dx}\ln(y) = \frac{d}{dx}\left(e^x \cdot \ln(x) \right)\\
\frac{1}{y} \frac{dy}{dx}= e^x \frac{d}{dx}\ln(x) + \ln(x) \frac{d}{dx}e^x \\
\frac{1}{y} \frac{dy}{dx}= e^x \frac{1}{x} + \ln(x) e^x \\
\frac{dy}{dx}= y \left( e^x \frac{1}{x} + \ln(x) e^x \right)
\]
using \( y=x^{e^x} \) we get
\[ \frac{dy}{dx}= x^{e^x} \left( e^x \frac{1}{x} + \ln(x) e^x \right) \]