Having trouble with exponents and differentiation

[Cadbury]

S=10(1−e^-0.4A )(1−e^-t )

A = 3, t = 3

Question asks, What is the effect of a small increase in A, dA, while keeping t constant?

Absolutely screwed with this one :P

[jemidiah]

You want to get an equation with dA in it. The most obvious way I can think of to do so is to take the derivative of S with respect to A. This will give you dS/dA = something, so then if you're given some small change in S (dS), you can approximate dA as dA = dS / something from before.

How do you take the derivative of the expression you have with respect to A? If you apply the product and chain rules rigorously, you don't need knowledge of partial derivatives, when you assume t is constant with respect to changes in A.

If you need more or this doesn't make sense, I'd be happy to clear it up.

[Cadbury]

Well the exercise is based on the topic of partial differentiation so I think that's the method I'm supposed to using, and if you would clear it up that'd be a great help, thanks.

[jemidiah]

Here's an example. Note that I'm ignoring writing del's instead of d's to signify partial derivatives.

Problem:
Given S = (1+A)*(2+t), find the partial derivative of S with respect to A. That is, take the derivative of S assuming t is constant.


Solution 1:
dS/dA = d[(1+A)*(2+t)]/dA = derivative of the first times the second + derivative of the second times the first = d(1+A)/dA * (2+t) + (1+A) * d(2+t)/dA = 1*(2+t) + (2+A)*0 = 2+t.

The zero comes from the fact that t does not change with changes in A--this is the assumption that t is constant--so that dt/dA = 0. Everything else is a straight application of the product rule.


Solution 2:
dS/dA = d[(1+A)*(2+t)]/dA = d[2+t+2A+At]/dA = d(2)/dA + dt/dA + d(2A)/dA + d(At)/dA = 0 + 0 + 2 + d(At)/dA.

What is d(At)/dA? Assuming t is just a constant, treat it as if it were a number like 2 or 3. Then d(At)/dA is clearly just t, so dS/dA = 2+t as above.

Alternatively, you can again apply the product rule rigorously to get d(At)/dA = dA/dA * t + A * dt/dA = 1*t + A*0 = t, which gives the same answer as before.



Note that in this case, S is implicity S(A, t), a function of two variables, so partial derivatives are needed, since regular derivatives are only defined on a function of one variable.

In your case you'll do basically this same process, though you'll need to remember the chain rule as well.

[MaximilianMayrhofer]

Whether the function increases or decreases for a small increase dA can be shown by the differential dS/dA. If it is positive, then the function increases, otherwise it decreases. Finding the differential we get,

dS/dA = (10/(1-e^-t))(0.4e^-0.4A)

Since 0.4e^-0.4A > 0 for all values of A, therefore the equation (10/(1-e^-t)) will determine the sign of the differential. Solving we get:

For t > 0, S(A + dA) > S(A) **
For t < 0, S(A + dA) < S(A) **

** Where dA > 0

There is no solution at t = 0.