# Differenntiate equation, help

#### dgamma1

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

#### AllanCuz

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

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

#### dgamma1

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!

#### AllanCuz

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 $$\displaystyle y^{ \prime }$$