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

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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Mar 9, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \tanh \frac{\ln x}{x} - \tan x$

#### 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;

$y = \ln \left(\cosh \left(\ln x\right)\right) + \ln \left(\cos x\right)$

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

Let's say $w = \cosh \left(\ln x\right)$
and $u = \cos x$
now;

$y = \ln \left(w\right) + \ln \left(u\right)$
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)

$\frac{d}{\mathrm{dx}} \cdot y = \frac{d}{\mathrm{dx}} \cdot \ln \left(w\right) + \frac{d}{\mathrm{dx}} \cdot \ln \left(u\right)$

Well, we know the derivative of $\ln x$ to be $\frac{1}{x}$ and using the chain rule we get;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{w} \cdot \frac{\mathrm{dw}}{\mathrm{dx}} + \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So let's go back to $u \mathmr{and} w$ and find their derivatives

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \cos x = - \sin x$
and
$\frac{\mathrm{dw}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \cosh \left(\ln x\right) = \sinh \left(\ln x\right) \cdot \frac{1}{x}$ (using the chain rule)

Plugging our newly found derivatives, and u, and w back into $\frac{\mathrm{dy}}{\mathrm{dx}}$ we get;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cosh} \left(\ln x\right) \cdot \sinh \frac{\ln x}{x} + \frac{1}{\cos} x \cdot - \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sinh \frac{\ln x}{x \cosh \left(\ln x\right)} - \sin \frac{x}{\cos} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \tanh \frac{\ln x}{x} - \tan x$

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

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