How do you find the derivative of #T=4cot^3(4z)#?

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Answer 1 · 2016-09-23T22:38:02

#T_z = -48 cot^2(4z) * csc^2 (4x)#

assuming you mean #(dT)/(dz)#, then by the product and chain rules:

#(dT)/(dz) = (d (4cot^3(4z)))/(dz)= 3*4 cot^2(4z) * d/dz(cot(4z))#

# = 12 cot^2(4z) * d/dz(cot(4z)) qquad triangle#

now #d/(dx) cot p = - csc^2 p #

if we have #p = 4 z# then by the Chain Rule

#d/(dz) cot p(z) = - csc^2 p (dp)/(dz) = = - csc^2 p * 4 = - 4 csc^2 z#

so #triangle# becomes

# = 12 cot^2(4z) * (- 4 csc^2 4z)#

# = -48 cot^2(4z) * csc^2 (4x)#

# = -48 (cos^2(4z))/(sin^2(4z)) * (1)/(sin^2 (4x))#

that simplification is going nowhere so we can stick with the previous answer

#T_z = -48 cot^2(4z) * csc^2 (4x)#