Here is our function:
f(x) = ln(sin x)^2 - 3x ln(sin x) + x^2 ln(cos^2 x^2)
Taking the derivative, we should have three smaller parts:
(df)/(dx) = (d/dx ln(sin x)^2) - (d/dx 3x ln(sin x)) + (d/dx x^2 ln(cos^2 x^2))
Let's label them:
g(x) = ln(sin x)^2
h(x) = 3x ln(sin x)
p(x) = x^2 ln(cos^2 x^2)
So it becomes:
(df)/(dx) = (dg)/(dx) - (dh)/(dx) + (dp)/(dx)
Let's find the derivatives one-by-one, starting from g(x). Let's... break this function composition down.
g_1 (x) = sin(x)
g_2 (x) = ln(x)
g_3 (x) = x^2
So g(x) = g_3 (g_2 (g_1 (x))). Let's take the derivative, one by one, using the chain rule:
dg_1 = d(sin(x)) = cos(x) dx
(dg_1)/(dx) = cos(x)
dg_2 = d(ln(g_1)) = (g_1)^-1 dg_1 = (sin(x))^-1 cos(x) dx
(dg_2)/(dx) = csc(x) cos(x)
dg_3 = d((g_2)^2) = 2g_2 dg_2 = 2 ln(g_1) csc(x) cos(x) dx
= 2 ln(sin(x)) csc(x) cos(x) dx
(dg_3)/(dx) = 2 ln(sin(x)) csc(x) cos(x)
Therefore (dg)/(dx) = 2 ln(sin(x)) csc(x) cos(x). Substitute that back:
(df)/(dx) = 2 ln(sin(x)) csc(x) cos(x) - (dh)/(dx) + (dp)/(dx)
Now, on to (dh)/(dx).
First using the product rule, we have that
dh = d(3x ln(sin x)) = 3x * d(ln(six(x))) + ln(sin(x)) * d(3x)
Then we need to find d(3x) and d(ln(sin(x))). The first one is easy:
(d(3x))/(dx) = 3 rarr d(3x) = 3 dx
Putting that back:
(dh) = 3x * d(ln(sin(x))) + 3 ln(sin(x)) dx
As for d(ln(sin(x)))... well, this happens to be the same as the dg_2 we calculated earlier, so:
(dh) = 3x csc(x) cos(x) dx + 3 ln(sin(x)) dx
(dh)/(dx) = 3x csc(x) cos(x) + 3 ln(sin(x))
Substituting this back:
(df)/(dx) = 2 ln(sin(x)) csc(x) cos(x) - 3x csc(x) cos(x) - 3 ln(sin(x)) + (dp)/(dx)
For (dp)/(dx), we'll use the product rule again:
dp = x^2 d(ln(cos^2 x^2)) + ln(cos^2 x^2) d(x^2)
Let's do d(x^2):
(d(x^2))/(dx) = 2x rarr d(x^2) = 2x dx
Put it back:
dp = x^2 d(ln(cos^2 x^2)) + 2x ln(cos^2 x^2) dx
Now we just need d(ln(cos^2 x^2)), which is more like d(ln((cos(x^2))^2)), and we'll label dq:
dp = x^2 dq + 2x ln((cos(x^2))^2) dx
Let's break down the composition of dq:
q_1 (x) = x^2
q_2 (x) = cos(x)
q_3 (x) = x^2
q_4 (x) = ln(x)
So q(x) = q_4(q_3(q_2(q_1(x)))). Yes, q_1 and q_3 are the same, but when peeling off the onion, we need different layers to be labeled differently.
(dq_1)/(dx) = 2x rarr dq_1 = 2x dx
dq_2 = d(cos(q_1)) = -sin(q_1) dq_1 = -sin(x^2) 2x dx
(dq_2)/(dx) = -sin(x^2) 2x
dq_3 = d((q_2)^2) = 2q_2 dq_2 = 2cos(q_1) * -sin(x^2) 2x dx
= -4x sin(x^2) cos(x^2) dx
(dq_3)/(dx) = -4x sin(x^2) cos(x^2)
dq_4 = d(ln(q_3)) = (q_3)^-1 dq_3
= (q_2)^-2 * -4x sin(x^2) cos(x^2) dx
= (cos(q_1))^-2 * -4x sin(x^2) cos(x^2) dx
= (cos(x^2))^-2 * -4x sin(x^2) cos(x^2) dx
= -4x sin(x^2) cos(x^2) (cos(x^2))^-2 dx
= -4x sin(x^2) (cos(x^2))^-1 dx
= -4x sin(x^2) sec(x^2) dx
(dq_4)/(dx) = -4x sin(x^2) sec(x^2)
So,
(dq)/(dx) = -4x sin(x^2) sec(x^2) rarr dq = -4x sin(x^2) sec(x^2) dx
Let's substitute this back into (dp)/(dx):
dp = x^2 * -4x sin(x^2) sec(x^2) dx + 2x ln((cos(x^2))^2) dx
dp = -4 x^3 sin(x^2) sec(x^2) dx + 2x ln((cos(x^2))^2) dx
(dp)/(dx) = -4 x^3 sin(x^2) sec(x^2) + 2x ln((cos(x^2))^2)
And finally this into (df)/(dx):
(df)/(dx) = 2 ln(sin(x)) csc(x) cos(x) - 3x csc(x) cos(x) - 3 ln(sin(x)) - 4 x^3 sin(x^2) sec(x^2) + 2x ln((cos(x^2))^2)
