Question

Is this correct use of the chain-rule? y = e^(lna . x) = (e^x)^lna
y' = lna . (e^x)^(lna - 1) . e^x = lna . (e^x)^(lna - 1 + 1) = lna . e^(lna . x)

Answer 1

I see no problem, but just to be sure, let's try to integrate lna.e^(x.lna). As lna it a constant, we can put it out of the integrand, so we calculate Int(e^(x.lna)), which, by advanced guessing, yields (e^(x.lna))/lna. Mutiplying by the original lna which was outside the integrand, we get e^(x.lna), out original function. So the derivation is correct.

Answer 2

Sorry, but no. Your variable needs to be in the base if you are going to use the power rule for derivatives. When your variable is in the exponent, you need to use different methods, such as the chain rule.
Your answer was correct, but the method was off.
Chain rule basically says to take the derivative of the outside function while leaving the inside alone. Then multiply by the derivative of the inside function.
\[d/dx [e ^{x \ln a}]=e ^{x \ln a}.d/dx[x \ln a]=e ^{x \ln a}.\ln a\]

Answer 3

He indeed applied chain rule, AFAIK. Calling \[f(x) = e^{x}\] we have \[d/dx[f(x)^{lna}] =f'(x).lna.f(x)^{lna-1} = e^{x}.lna.(e^{x})^{lna-1}\] \[=lna.(e^{x})^{lna-1+1}=lna.e^{xlna}\]

it is just another form of the chain rule.

Answer 4

It was definitely an application of the chain rule. I don't believe it was a good use of the chain rule. I would say it is questionable if it was a "correct use" of the chain rule.

It was an elaborate work around. My question would be: why would you do this?

Answer 5

Tx, I was just looking at different ways to solve d/dx[e^xlna ]