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

Answer 1

To find the derivatives I always remind people to take note of the layers (we need the derivatives of each layer to properly determine the derivative using the chain rule)

y = csc[sec{x -8 (³√(3x))}] = csc [sec q]

Think p = sec q and q = x -8 (³√(3x)) also note a cube root is the same as one-third power

so this function has functions of type csc [p], sec {q}, and power functions contained in it. I used different brackets to show the bounds of different functions being composed. The derivative of csc p is - csc p cot p

The derivative of sec q is sec q tan q , the derivative of q = x -8 (³√(3x)) is x -8 (³√(3x))

Now we can put the parts of the chain rule together.

If y = csc [sec q] then y' = -csc [sec q] cot [sec q] (sec q) (tan q) .q'(x)

=-csc[sec {x -8 (³√(3x))] cot[sec {x -8 (³√(3x)) ] sec {x -8 (³√(3x)) } tan {x -8 (³√(3x)) } [x -8 (³√(3x))]