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

1 Answer
Mar 9, 2018

Answer:

#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 :)