How do you differentiate #f(x)=(ln(sinx)^2-3xln(sinx)+x^2ln(cos^2x^2)# using the chain rule?

Question

Emilia Aguilar · Accepted Answer

Take the derivative of each term, using the product rule and chain rule, to get

#(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)#

which may be factored as needed.

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)#