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

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

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