Solve for dy/dx given y = csc (sec(x - 8 ∛3x).

Answer 1

Start from the two trigonometric derivatives

1) Derivative of csc(x)=-csc(x)cot(x)

2) Derivative of sec(x)=-sec(x)tan(x)

Hence, by applying twice the chain rule

d/dx(F(G(x)))=F'(G(x·))G'(x)

you have

dy/dx=-csc(sec(x-8·3^1/3·x^1/3))·cot(sec(x-8·3^1/3·x^1/3) · sec(x-8·3^1/3·x^1/3) · tan(x-8·3^1/3·x^1/3) · (1-8·3^1/3·(1/3·)x^-2/3)

where I 3^1/3 and x^1/3 denote the cubic root of 3 and x, respectively.

Best,

Davide