Differrentiate this function

**Redeemer_Pie** · October 15th 2010, 09:45 AM

Hi all, just stuck with a simple equation at the moment and dont really know how to solve it. Hope you guys could assist me! Thanks.

The function is given as:

y= 1/C (1 - e^(-C(z - x))

what is the answer given i have differrentiate: dy/dx

C is a scalar constant

Thanks and much appreciated!

**e^(i*pi)** · October 15th 2010, 10:40 AM

Assuming you mean: $\dfrac{1}{C}\left(1-e^{-C(z - x)}\right)$ and that Z and C are constants.

Expand to give $\dfrac{1}{C}\left(1-e^{Cz-Cx}\right) = \dfrac{1}{C} - \dfrac{1}{C}\left(e^{Cz-Cx}\right)$

Using the laws of exponents we can write $e^{Cz-Cx} = e^{Cz} \cdot e^{-Cx}$

Group the constants: $\dfrac{1}{C} - \dfrac{e^{Cz}}{C}\,e^{-Cx}$

That should be quite easy to differentiate using the chain rule

**Redeemer_Pie** · October 15th 2010, 05:14 PM

thanks^

But im still confused.

How come the expotential became Cz-Cx not:

-Cz + Cx?

**Redeemer_Pie** · October 15th 2010, 05:50 PM

Can someone give me the final result of the equation?

Im not too sure whether the constant disappears

Thanks

**Educated** · October 15th 2010, 06:37 PM

As you said before, C is a scalar constant so it can be either positive or negative.

You won't learn anything if we give you the answer.

$y= \dfrac{\left(1-e^{Cz - Cx}\right)}{C}$

$y= \dfrac{1}{C} - \dfrac{e^{Cz - Cx}}{C}$

$y= \dfrac{1}{C} - \dfrac{e^{Cz} \cdot e^{-Cx}}{C}$

Can you differentiate from here?
And no, the constants do not disappear.

**Redeemer_Pie** · October 15th 2010, 06:52 PM

i thought the answer will be like:

1/C - e^Cz . e^-Cx

**HallsofIvy** · October 16th 2010, 04:54 AM

C is a constant and so is $\frac{1}{C}$ . Its derivative is 0. As Educated told you,
$y= \frac{1}{C}- \frac{1}{C}(e^{Cz}e^{-Cx}$

The derivative of the first $\frac{1}{C}$ is 0.

What is the derivative of $Ae^{-Cx}$ where A and C are constants?

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