Parametric differentiation with 3 variables at a given value

Question

I am stumped by this question:

Find the derivative of $y=e^u$ where $u=5v^3-5$ and $v=1+\ln x^2$ when $x=1$

I tried subbing in the value for $x$ to get $v=1$ then putting that into the equation for $u$ to get $u=0$ then putting that into the derivative of y to get $y'=1$ but that's not right and I'm unsure of what else to try.

Answer 1

You cannot do the substitution before differentiation. Otherwise, everything becomes constant and the derivative is $0$.

\begin{align*} \frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}\\ &=e^u\cdot(15v^2)\cdot\left(\frac{2}{x}\right) \end{align*}

When $x=1$, $v=1$ and $u=0$

So, $\displaystyle \left.\frac{dy}{dx}\right|_{x=1}=e^0(15)(2)=30$.