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