How can I differentiate this equation? 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?
 A: 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
$$
y'(x)=b(ae^{ax}-(-a)e^{-ax})=ab(e^{ax}+e^{-ax})
$$
A: 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
$$\frac{\mathrm d}{\mathrm dx} b(e^{ax} - e^{-ax}) = b(ae^{ax} - (-a)e^{-ax}) = ab(e^{ax}+e^{ax})$$
A: First Destribute the b.
$be^{ax}-be^{-ax}$
Then differentiate:
$abe^{ax}+abe^{-ax}$
A: $$
y = b\left(e^{ax}-e^{-ax}\right) $$
Here are the steps
$$
\frac{d}{dx}\left[b\left(e^{ax}-e^{-ax}\right)\right]=b \frac{d}{dx}\left[e^{ax}-e^{-ax}\right] $$
$$ =b \left(\frac{d}{dx}\left[e^{ax}\right]-\frac{d}{dx}[e^{-ax}]\right)
$$
$$= b\left(e^{ax}\frac{d}{dx}[ax]-e^{-ax}\frac{d}{dx}[-ax]\right) 
$$
$$= b\left(ae^{ax}\frac{d}{dx}[x]+ae^{-ax}\frac{d}{dx}[x]\right)$$
$$= b\left(ae^{ax}+ae^{-ax}\right)$$
$$= ab\left(e^{ax}+e^{-ax}\right) $$
