Hi everybody
I am having problems with the following:
1. Find dy/dx for y=(arccosh sqrt X)^3x
2. Find integral of (sinhx.cosh2x)dx
Any help will be much appreciated.
Thanks
Lets start with number 1.
First you need the derivative of $\displaystyle (f(x))^{3x}$.
We start by
$\displaystyle
\displaystyle y = f(x)^{3x}$
$\displaystyle \displaystyle \ln(y) = 3x\ln(f(x))$
Next we take the derivative of both sides to get
$\displaystyle \displaystyle \frac{1}{y}\frac{dy}{dx} = 3 \ln(f(x)) + 3x\frac{1}{f(x)}f'(x)$
$\displaystyle \displaystyle \frac{dy}{dx} = 3 y \left(\ln(f(x))+3x\frac{1}{f(x)}f'(x)\right) = 3 f(x)^{3x} \left(\ln(f(x))+3x\frac{1}{f(x)}f'(x)\right) $ (1)
In your case $\displaystyle f(x) = \arccosh(\sqrt{x}) = \ln(\sqrt{x}+\sqrt{x-1})$
The derivative of that is
$\displaystyle \displaystyle f'(x) = \frac{1}{\sqrt{x}+\sqrt{x-1}}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x-1}}\right)$
What you now need to do is plug this in in expression (1)