Solving Derivative of ln[cos(w-1)]: Need Help!

[dec1ble]

I am presenting a problem in front of the class tomorrow and I am slightly confused on the steps for my problem. The problem is:

Find the derivative of the given function

f(w) = ln[cos(w-1)]

The answer in the back of my book shows the derivative is -tan(w-1) - but I'm my steps aren't giving that answer - could anyone show me the steps to use in order to get that answer? Thanks a lot!

 

[saltydog]

I am presenting a problem in front of the class tomorrow and I am slightly confused on the steps for my problem. The problem is:

Find the derivative of the given function

f(w) = ln[cos(w-1)]

The answer in the back of my book shows the derivative is -tan(w-1) - but I'm my steps aren't giving that answer - could anyone show me the steps to use in order to get that answer? Thanks a lot!

Break it down:

Suppose have a function of x as u(x) and have the expression:

ln[u(x)]

Would you not, using the chain rule, take the derivative of the logarithm which is $\frac{1}{u(x)}$, then take the derivative of u(x) to get:

$$\frac{1}{u(x)}u^{'}$$

Same dif with the cosine of a function right? Derivative is minus the sine of the function times the derivative of the function. In your case, it's nested three times (well w-1 is a function of w in which the derivative is just 1).

 

[dec1ble]

if i have $$\frac{d}{dx} \ln u = \frac{1}{u}$$

then i find d/dx cos(w-1) = -sin(1)

then plug that into the u?

so 1/-sin(1) - then I am stuck