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

D/dx (2cos(x)sin❻¹(x)) right, 2cos(x)sin❻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = you d/dx(v) + v d/dx(u) you = 2cos(x) v = sin❻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin❻¹(x)) we'll set y=sin❻¹(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)sin❻¹(x)) 2cos(x)/sqrt(1-x²) - 2sin(x)sin❻¹(x).

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