Derivatives of trigonometric functions - Question

[Joe_K]

Homework Statement

Find the Derivative of:

(ln(cos4x)) / 12x^2

Homework Equations

y' ln(x) = 1/x

The Attempt at a Solution

I have determined the correct answer, but I am still confused as to how I came to the solution. Starting with the numerator, the derivative of cos is -sin. This would produce ln(-4sin(4x)) in the numerator. Now, to take the derivative of the natural log, I would do ln(x)= 1/x, correct? So 1/-4sin(4x)?

Why does the final numerator come out to -4tan(4x)? I know I am missing a basic rule or overlooking something... but why does it become tangent?

Thank you

 

[Ray Vickson]

Homework Statement

Find the Derivative of:

(ln(cos4x)) / 12x^2

Homework Equations

y' ln(x) = 1/x

The Attempt at a Solution

I have determined the correct answer, but I am still confused as to how I came to the solution. Starting with the numerator, the derivative of cos is -sin. This would produce ln(-4sin(4x)) in the numerator. Now, to take the derivative of the natural log, I would do ln(x)= 1/x, correct? So 1/-4sin(4x)?

Why does the final numerator come out to -4tan(4x)? I know I am missing a basic rule or overlooking something... but why does it become tangent?

Thank you

You would NOT get ln(-4sin(4x)) in the numerator. You are forgetting the Chain Rule: [f(g(x))]' = f '(g(x) )* g '(x). Apply this to f(.) = ln (.) and g(x) = cos(4x).

RGV

 

[Mentallic]

Find the Derivative of:

(ln(cos4x)) / 12x^2

First of all, this is a quotient so you need to use the quotient rule.

$$y=\frac{u}{v}$$ where u and v are functions of x, then $$y'=\frac{v'u-u'v}{v^2}$$

Now, only tricky part there would be to determine u'. Since u=ln(cos(4x)) then we need to apply the chain rule, and Ray Vickson has already shown how this should be done.