How to differentiate functions with x in their exponents

[canger]

I'm not sure how to differentiate y with respect to x for:

y = (7x^2)^[x^2]

Any ideas?

 

[Isak BM]

y = {(7x^2)}^{x^2}

Mission: To find \frac{dy}{dx}

We want to eliminate the problem of the exponent x^2.
There are several ways, but here's one neat way. Take the natural logarithm of both sides. In each step i will write in red what mathematical identity I have used to get to the expression.

\ln(y) = \ln({(7x^2)}^{x^2})


\ln(y) = x^2\ln(7x^2) \ln{(a^b)} = b\ln(a)

\ln(y) = x^2\ln{({(\sqrt{7}x)}^2)} ab^2 = {(\sqrt{a}b)}^2

\ln(y) = 2x^2\ln{(\sqrt{7}x)} \ln{(a^b)} = b\ln{a}

Now raise e to the power of each side to get.

y = e^{2x^2\ln(\sqrt{7}x)}

What remains is just deriving this expression, and to do so you only need to know the chain rule and product rule, and how to derive e^x.

Ok, so let's do the remaining.

\frac{dy}{dx} = \frac{d}{dx}e^{2x^2\ln(\sqrt{7}x)} = e^{2x^2\ln(\sqrt{7}x)}\cdot \frac{d}{dx}2x^2\ln(\sqrt{7}x) Chain rule

Now

\frac{d}{dx}2x^2\ln(\sqrt{7}x) = 4x\ln(\sqrt{7}x) + 2x^2 \frac{1}{x} = 2x(\ln(7x^2) +1) Product rule

Thus

\frac{dy}{dx} = 2x(\ln(7x^2) +1)e^{2x^2\ln(\sqrt{7}x)} = 2x(\ln(7x^2) +1){(7x^2)}^{x^2}

 

[DivisionByZro]

Use the chain rule with u=x^2, and v=x^2. Factor out the constant, and your answer should appear quickly.