Expert: Scotto - 2/3/2011

Question
In trying to refresh for a course in advanced math going back to my calc 2 Taylor Series math.  However, I appear a bit rusty.  I can do simple lineaizations of functions, but not more complicated ones.  For example, take the function:

(3 +5ln(1 + 4sin2x))^(1/2)   

Just so its understood, (3 +5ln(1 + 4sin2x)) is UNDER a square root.

I want to linearize this for SMALL x.

I want to do this step-by-step linearization.

Here’s what I do so far:  

Sin2x ~ x

But I get confused now what to do with the ln(1+x), and ultimately what I do afterwards to get at least the linearized function.  The Taylor series part of it is easy.

Hope you can help.

Answer
To take the derivative of this function, the chain rule is involved repeated.
First, we have the function as √(3 + 5ln(2 + 4sin(2x))), correct?

To take the derivative, it is the derivative of a squareroot first.

It is known that √f(x) is f'(x)/(2√f(x)).  For this problem, f(x) = 5*ln(1 + 4sin(2x)).

It is known that if we have f(x) = K*ln(g(x)), then f'(x) = K*g'(x)/g(x).
In this case, g(x) = 1 + 4sin(h(x)), so g'(x) = 4cos(h(x))h'(x).

Next, we note that h(x) = 2x, and so h'(x) = 2.

Putting this back in g'(x) gives g'(x) =  4cos(2x)2 = 8(cos(2x)).
Putting g'(x) and g(x) back into f'(x) gives f'(x) with K as 5 gives
f'(x) = 5(8(cos(2x)))/(1 + 4sin(h(x))).

To find the derivative of the function, put back f(x) and f'(x) into f'(x)/(2√f(x)).
This gives [5(8(cos(2x)))/(1 + 4sin(2x))]/[2(5*ln(1 + 4sin(2x)))].
This is the same as [40(cos(2x))/[10(1 + 4sin(2x))ln(1 + 4sin(2x)))].

Note that 40/10 = 4, so this is the same as [4(cos(2x))/[(1 + 4sin(2x))ln(1 + 4sin(2x)))].
Basically, if we say m(x) = sin(2x), then m'(x) = 2cos(2x), so the derivative can be
rewritten 2m'(x)/[(1+4m(x))ln(1+4m(x))].

That is the first derivative, and you say the rest can be gotten once you have that...