Question

y-sin^-1(2x^3) =cos^-1(2x^3) find dy/dx​

Answer 1

Answer:d/dx[Sin^-1 (x)] = 1/Square Root [ 1- x^2]

Let

y=Sin^-1 (2x)

dy/dx= d/dx [Sin^-1 (2x)]

dy/dx= 1/Square Root [ 1- (2x)^2]

dy/dx= 1/Square Root [ 1- 4x^2]

And

y=Sin^-1 (2x-3)

dy/dx= d/dx [Sin^-1 (2x+3)]

dy/dx= 1/Square Root [ 1- (2x+3)^2]

dy/dx= 1/Square Root [ 1- {4x^2+12x+9}]

dy/dx= 1/Square Root [ 1- 4x^2 - 12x - 9]

dy/dx= 1/Square Root [ - 4x^2 - 12x - 8]

dy/dx= 1/Square Root [4{ - x^2 - 3x - 2}]

dy/dx= 1/2 Square Root [ - x^2 - 3x - 2]

Step-by-step explanation: