Hello, Can anyone please help me to differentiate \(\displaystyle

e^{(10.234 - 6.984*0.96^{(t+3.3)})}* 1000 * 0.96^t\)

Any help would be appreciated.

Thanks

Let's make it look a little nicer by bringing all the constants out front.

\(\displaystyle 1000 e^{10.234} [.96^t][e^{(-6.984).96^{t+3.3}} ] \)

Leave the constants for now

\(\displaystyle [.96^t][e^{(-6.984).96^{t+3.3}} ] \)

\(\displaystyle [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] \)

Let \(\displaystyle y = [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] \)

\(\displaystyle lny = ln [.96^t] + ln[e^{(-6.984)(.96^{3.3}).96^{t}} ] \)

\(\displaystyle lny = t ln(.96) + (-6.984)(.96^{3.3}).96^{t} ln(e) \)

\(\displaystyle \frac{1}{y} y^{ \prime} = ln(.96) - (6.984)(.96^{3.3})[.96^{t}]^{\prime} \)

To find the unknown derivative on the right hand side

Let \(\displaystyle p = .96^t \)

\(\displaystyle lnp = t ln .96 \)

\(\displaystyle \frac{1}{p} p^{ \prime } = ln .96 \)

\(\displaystyle p^{ \prime } = .96^t ln .96 \)

Subbing back in

\(\displaystyle \frac{1}{y} y^{ \prime} = ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] \)

Finally,

\(\displaystyle y^{ \prime} = [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] \)

But remember we have that constant out front, so in total we get

\(\displaystyle y^{ \prime} = 1000 e^{10.234} [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] \)