I've attached a write-out of the question and my answer.

~Declan Bolster

I'm taking a more closer look now and there's a few things I should comment on from my first glance:

- You're already using \(a\) and \(b\) for constants already. It's not recommended to now define stuff like \(a(x)\) \(b(x)\), where you use the same letter.

- Your derivative is in terms of \(t\). Use functions in terms of \(t\), not \(x\).

- \( 1 - (be)^{dt}\) isn't the same thing as \( (1-be^t)^d\).

So I feel the problem got made a bit more convoluted than what it had to be! This is how I would've done it. First rewrite:

\[ y = 1000 a^d c (be)^t \left( 1- (be)^t\right)^d. \]

Since \(a\), \(c\) and \(d\) are constants here, we can just leave the \(1000 a^d c\) bit in front, and focus on everything else for the product rule.

Set up for the product rule:

\begin{align*}

f(t) &= (be)^t \\

g(t) &= \left( 1 - (be)^t\right)^d.

\end{align*}

Judging by your working out, I'm assuming you're allowed to use the rule \( \boxed{\frac{d}{dx}a^x = a^x \ln (a)} \). Applying it here, we have

\begin{align*}

f^\prime (t) &= (be)^t \ln (be)\\

g^\prime(t) &= (be)^t\ln (be) \times d\left(1 - (be)^t \right)^{d-1}

\end{align*}

having used the chain rule on the second computation.

Note that \(\ln(be) = \ln(b)+\ln(e) = \ln(b)+1\), which does not conflict with some of the stuff I see you did there.

Hence, putting everything together, the derivative is

\begin{align*}

\frac{dy}{dt} &= 1000 a^d c \left[g(t)f^\prime(t) + f(t)g^\prime(t) \right]\\

&= 1000 a^d c\left[ \left(1 - (be)^t\right)^d (be)^t \ln (be) + (be)^t (be)^t \ln (be) d \left(1-(be)^t \right)^{d-1} \right]

\end{align*}

This certainly can be simplified further, but it's still going to be ugly for sure. If you want me to simplify it, feel free to ask