How can I differentiate this equation?

Question

I need to differentiate this:

$$ y = b(e^{ax}-e^{-ax}) $$

I've got the solution from a book, but I don't found the process to differentiate it. The solution is:

$$ y = ab(e^{ax}+e^{-ax}) $$

Here is my own process:

$$ y = b(e^{ax}*(1-\frac{e^{-ax}}{e^{ax}}) $$

$$ ln(y) = ln(b)+ax+ln(1-\frac{e^{-ax}}{e^{ax}}) $$

$$ \frac{1}{y}\frac{dy}{dx} = a+\frac{1}{1-\frac{e^{-ax}}{e^{ax}}} = a+1-e^{2ax} $$

$$ \frac{dy}{dx} = (a+1-e^{2ax})*(b(e^{ax}-e^{-ax})) $$

Can someone help me, please?

Answer 1

No need to factor.

The derivative of a sum is the sum of the derivatives. Do you know what the derivative of $e^x$ is? And the Chain Rule? If so just apply them to get $$ y'(x)=b(ae^{ax}-(-a)e^{-ax})=ab(e^{ax}+e^{-ax}) $$

Answer 2

You're doing it much too complicated. $$\frac{\mathrm d}{\mathrm dx} e^x = e^x$$ So by the chain rule $$\frac{\mathrm d}{\mathrm dx} e^{ax} = ae^{ax}\\ \frac{\mathrm d}{\mathrm dx} e^{-ax} = -ae^{ax}$$ And thus by linearity of the differential $$\frac{\mathrm d}{\mathrm dx} b(e^{ax} - e^{-ax}) = b(ae^{ax} - (-a)e^{-ax}) = ab(e^{ax}+e^{ax})$$

Answer 3

First Destribute the b.

$be^{ax}-be^{-ax}$

Then differentiate:

$abe^{ax}+abe^{-ax}$

Answer 4

$$ y = b\left(e^{ax}-e^{-ax}\right) $$ Here are the steps $$ \frac{d}{dx}\left[b\left(e^{ax}-e^{-ax}\right)\right]=b \frac{d}{dx}\left[e^{ax}-e^{-ax}\right] $$ $$ =b \left(\frac{d}{dx}\left[e^{ax}\right]-\frac{d}{dx}[e^{-ax}]\right) $$ $$= b\left(e^{ax}\frac{d}{dx}[ax]-e^{-ax}\frac{d}{dx}[-ax]\right) $$ $$= b\left(ae^{ax}\frac{d}{dx}[x]+ae^{-ax}\frac{d}{dx}[x]\right)$$ $$= b\left(ae^{ax}+ae^{-ax}\right)$$ $$= ab\left(e^{ax}+e^{-ax}\right) $$