Finding derivative question

[rainyrabbit]

Finding derivative question please...

How should I find the derivative of y = x^(e^x)?

I tried using the chain rule along with the power rule, coming out to:
(e^x) (e^x) (X^(e^x - 1))

If I had took the natural log of both sides and then used implicit differentiation, I would have gotten as a derivative:
(x^(e^x)) (e^x) (1/x + ln x)
which is the correct answer according to my TI89.

Why wouldn't the first method work? Or was there any flaw?

By the way I just started Calculus as a high schooler.

 

[courtrigrad]

The power rule only works for functions of the form [tex] f(x) = x^{n} [/tex]. So you can't use it for functions in the form of [tex] f(x) = x^{g(x)} [/tex] You could use implicit differentiation. Or you could do the following:

[tex] x^{e^{x}} = (e^{\ln x})^{e^{x} [/tex].

[tex] \frac{d}{dx} ( x^{e^{x}})= \frac{d}{dx}(e^{\ln x}^{e^{x}}) = e^{\ln xe^{x}}\frac{d}{dx}(e^{x}\ln x) [/tex]

[tex] \frac{dy}{dx} = e^{\ln x e^{x}}(\frac{e^{x}}{x}+e^{x}\ln x) [/tex]

[tex] \frac{dy}{dx} = x^{e^{x}}e^{x}(\frac{1}{x}+ \ln x) [/tex]

Note: [tex] e^{\ln x e^{x}} = (e^{\ln x})^{e^{x}} = x^{e^{x}} [/tex]

 

[tacman]

Great job on using both the chain rule and power rule to find the derivative! You are on the right track. The first method you mentioned, using the chain rule and power rule, is the correct way to find the derivative of y = x^(e^x). However, there is a slight mistake in your calculation. The correct derivative should be (e^x) (x^(e^x - 1)) + (x^(e^x)) (e^x) (1 + ln x). Notice that the first term in parentheses should be (e^x), not (e^x)^2. This mistake is most likely due to a typo or miscalculation, so make sure to double check your work next time.

The second method you mentioned, using implicit differentiation, is also a valid way to find the derivative. However, it is a more advanced method and requires a good understanding of logarithmic and exponential functions. It is great that you are already exploring different techniques to solve problems in Calculus as a high school student. Keep up the good work and don't be afraid to ask for help when needed. Good luck in your studies!