The function y = sin^?1(3x + 1) is a composition, and so we must use the Chain Rule, given below, to find the derivative. d dx [f(g(x))] = f '(g(x))g'(x) For the given function sin^?1(3x + 1), the "inside" function is 3x + 1 and the "outside" function is f(x) = arcsin(x). Recall that the derivative of y = sin?1(x) is y' =?

dy/dx = 3/√1-(3x+1)²

Step-by-step exxplanation:

Given the inverse function y = sin^-1(3x+1), to find the derivative of the expression, we will use the chain rule as shown;

Let u = 3x+1 ...1

y = sin⁻¹u ...2

From equation 1, du/dx = 3

from equation 2;

Taking the sin of both sides;

siny = sin(sin⁻¹u)

siny = u

u = siny

du/dy = cosy

dy/du = 1/cosy

from trig identity, cos y = √1-sin²y

dy/du = 1/√1-sin²y

Ssince u = siny

dy/du = 1/√1-u²

According to chain rule, dy/dx = dy/dy*du/dx

dy/dx = 1/√1-u² * 3

dy/dx = 3/√1-u²

Substituting u = 3x+1 into the final equation, we will have;

dy/dx = 3/√1-(3x+1)²

answer: 5100 just subtract

step-by-step explanation: