Differenntiate equation, help

May 2010
2
0
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
 
Apr 2010
384
153
Canada
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


\(\displaystyle y^{ \prime} = 1000 e^{10.234} [.96^t][e^{(-6.984)(.96^{3.3}).96^{t}} ] ln(.96) - (6.984)(.96^{3.3})[.96^t ln .96 ] \)
 
May 2010
2
0
wow thanks for that. although there is a step I don't understand:


1. going from here:

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

to here: what did you do with the t in front of the Ln, and where did the Ln go from the left side.

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

Thanks for all your help so far!
 
Apr 2010
384
153
Canada
wow thanks for that. although there is a step I understand:


. going from here:

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

to here: what did you do with the t in front of the Ln, and where did the Ln go from the left side.

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

Thanks again for all your help!
No problem mate.

This is known as implicit differentiation.

Let's say we have

\(\displaystyle y = lnt \)

The derivative of \(\displaystyle lnt \) with respect to t is \(\displaystyle \frac{1}{t} t` = \frac{1}{t} (1) \)

In this case, we have

\(\displaystyle lny \) and the derivative of which (with respect to t) is \(\displaystyle \frac{1}{y} y^{ \prime } \)

In this case, however, y depends on t so \(\displaystyle y^{ \prime } \) doesn't equate to 1 like in the above example. So we must keep it.

By the same token,

\(\displaystyle tln(.96) \)

We are taking the derivative of the above with respect to t, which is equal to

\(\displaystyle ln(.96) \)

So basically we're taking the derivative of the left side with respect to t, and of the right side with respect to t. But since the left side is depdent on t and we dont have an equivilent expression for it, we leave it as \(\displaystyle y^{ \prime } \)