Differenntiate equation, help

**dgamma1** · May 23rd 2010, 04:40 AM

Hello, Can anyone please help me to differentiate $ e^{(10.234 - 6.984*0.96^{(t+3.3)})}* 1000 * 0.96^t$

Any help would be appreciated.

Thanks

**AllanCuz** · May 23rd 2010, 07:32 AM

Originally Posted by dgamma1

Hello, Can anyone please help me to differentiate $ 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.

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

Leave the constants for now

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

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

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

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

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

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

To find the unknown derivative on the right hand side

Let $p = .96^t$

$lnp = t ln .96$

$\frac{1}{p} p^{ \prime } = ln .96$

$p^{ \prime } = .96^t ln .96$

Subbing back in

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

Finally,

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

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

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

**dgamma1** · May 23rd 2010, 01:41 PM

wow thanks for that. although there is a step I don't understand:

1. going from here:

$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.

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

Thanks for all your help so far!

**AllanCuz** · May 23rd 2010, 01:51 PM

Originally Posted by dgamma1

wow thanks for that. although there is a step I understand:

. going from here:

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

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

$\frac{1}{y} y^{ \prime} = ln(.96) -$ $ 

Thanks again for all your help!

No problem mate.

This is known as implicit differentiation.

Let's say we have

$y = lnt$

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

In this case, we have

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

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

By the same token,

$tln(.96)$

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

$ln(.96)$

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 $y^{ \prime }$

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