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 ] $
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!
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 } $