Take sin of both sides:

sin(y) = 4x + 1

Differentiate implicitly:

cos(y) * y' = 4

Divide both sides by cos(y):

y' = 4 / cos(y)

To get it back into terms of x, draw up a right-angled triangle labeling the opposite side (4x + 1), the hypotenuse 1 and the adjacent √[1 - (4x + 1)^2] using Pythagoras Theorem and find cos of that angle:

y' = 4 / √[1 - (4x + 1)^2]

y' = 4 / √(-16x^2 - 8x)

Brittany - the derivative of sin^-1(x) is 1/sqrt(1 - x^2) so all we have to do is apply the Chain Rule at the end i.e. If y = sin^-1(4x + 1) then dy/dx = {1/sqrt[1 - (4x)^2)]}*d/dx(4x + 1) = 4/sqrt(1 - 16 x^2), and that's as far as you can take it.

Let u = 4x+1

Then du/dx = 4

dy/dx = 1/sqrt(1-u^2) X du/dx

= 4/sqrt(1 - (4x+1)^2)

= 4/sqrt(-16x^2 - 8x)

y = arcsin(4x + 1)

sin(y) = 4x + 1

Derive implicitly, then solve for dy/dx

cos(y) * dy = 4 * dx

dy/dx = 4 / cos(y)

dy/dx = 4 / sqrt(1 - sin(y)^2)

dy/dx = 4 / sqrt(1 - (4x + 1)^2)

dy/dx = 4 / sqrt(1 - 16x^2 - 8x - 1)

dy/dx = 4 / sqrt(-16x^2 - 8x)

dy/dx = 4 / sqrt((-8x) * (2x + 1))

This link has a step by step solution to your problem:

http://www.symbolab.com/solutions/derivatives?quer...

Hope this helps