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.