1. ## Differentiating yields different answers

Hi,

I have the following:
$y = Ate^{-2x}$
$y ^{\prime} = Ae^{-2x}-2Ate^{-2x}$

For $y^{\prime\prime}$ i seem to get different answers, depending on the method.

$\frac{d}{dx}f(x)\cdot g(x) = f^{\prime}(x)\cdot g(x) + f(x)\cdot g^{\prime}(x)$

1. $\frac{d}{dx}~Ae^{-2x}-2Ate^{-2x} = -2Ae^{-2x} + 4Ae^{-2x}$
2. If i factor out the e term. $\frac{d}{dx}~e^{-2x}\cdot (A-2At) = -2e^{-2x} \cdot (A-2At) + e^{-2x}\cdot -2A = -4Ae^{-2x} + 4Ate^{-2x}$

One of these is more correct than the other, which is it?

2. Originally Posted by Jones
Hi,

I have the following:
$y = Ate^{-2x}$
$y ^{\prime} = Ae^{-2x}-2Ate^{-2x}$

For $y^{\prime\prime}$ i seem to get different answers, depending on the method.

$\frac{d}{dx}f(x)\cdot g(x) = f^{\prime}(x)\cdot g(x) + f(x)\cdot g^{\prime}(x)$

1. $\frac{d}{dx}~Ae^{-2x}-2Ate^{-2x} = -2Ae^{-2x} + 4Ae^{-2x}$
2. If i factor out the e term. $\frac{d}{dx}~e^{-2x}\cdot (A-2At) = -2e^{-2x} \cdot (A-2At) + e^{-2x}\cdot -2A = -4Ae^{-2x} + 4Ate^{-2x}$

One of these is more correct than the other, which is it?
You are mixing the x and t when calculating the derivative.
If the power of "e" is -2t, then you have a product of 2 functions of "t".
If you differentiate with respect to x, you do not know what the derivative of "t" is, if t is itself a function of x.

3. Originally Posted by Jones
Hi,

I have the following:
$y = Ate^{-2x}$
$y ^{\prime} = Ae^{-2x}-2Ate^{-2x}$

For $y^{\prime\prime}$ i seem to get different answers, depending on the method.

$\frac{d}{dx}f(x)\cdot g(x) = f^{\prime}(x)\cdot g(x) + f(x)\cdot g^{\prime}(x)$

1. $\frac{d}{dx}~Ae^{-2x}-2Ate^{-2x} = -2Ae^{-2x} + 4Ae^{-2x}$
2. If i factor out the e term. $\frac{d}{dx}~e^{-2x}\cdot (A-2At) = -2e^{-2x} \cdot (A-2At) + e^{-2x}\cdot -2A = -4Ae^{-2x} + 4Ate^{-2x}$

One of these is more correct than the other, which is it?
The mistake you are doing, is that you are differentiating with respect to t sometimes and x the others. $\frac{d}{dx}$ means you are differentiating with respect to x, so, $\frac{d}{dx}At = A\frac{dt}{dx}$. What you didn't notice is that $\frac{d}{dx}At \neq A$
Hope that helps