Question

How to differentiate amd simplify: ln(cosh(ln x) cos(x)) ?

Answer 1

`dy/dx = tanh(lnx)/x - tanx`

Explanation:

I like to set the problem equal to y if it is not already. Also it will help our case to rewrite the problem using properties of logarithms;

`y = ln(cosh(lnx)) + ln(cosx)`

Now we do two substitutions to make the problem easier to read;

Let's say `w = cosh(lnx)`
and `u = cosx`
now;

`y = ln(w) + ln(u)`
ahh, we can work with this :)

Let's take the derivative with respect to x of both sides. (Since none of our variables are x this will be implicit differentiation)

`d/dx*y = d/dx*ln(w) + d/dx*ln(u)`

Well, we know the derivative of `lnx` to be `1/x` and using the chain rule we get;

`dy/dx = 1/w * (dw)/dx + 1/u*(du)/dx`

So let's go back to `u and w` and find their derivatives

`(du)/dx = d/dxcosx = -sinx`
and
`(dw)/dx = d/dxcosh(lnx) = sinh(lnx)*1/x` (using the chain rule)

Plugging our newly found derivatives, and u, and w back into `dy/dx` we get;

`dy/dx = 1/cosh(lnx) * sinh(lnx)/x + 1/cosx*-sinx`

`dy/dx = sinh(lnx)/(xcosh(lnx)) - sinx/cosx`

`dy/dx = tanh(lnx)/x - tanx`

If this can be simplified further, I haven't learned how. I hope this helped :)