What is the derivative of 2cosx times the inverse of sinx?

Answer 1

d/dx (2cos(x)sin⁻¹(x))

right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule:

d/dx(uv) = u d/dx(v) + v d/dx(u)

u = 2cos(x)

v = sin⁻¹(x)

d/dx(u) = -2sin(x)

to find d/dx(sin⁻¹(x)) we'll set

y=sin⁻¹(x)

x=sin(y)

dx/dy = cos(y)

dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1,

dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)]

so plugging all this into our product rule,

d/dx (2cos(x)sin⁻¹(x))

= 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).