The Derivative of arccosx
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fantastic and easy to understand. thanks!
garretonzo16 4 months ago
If you could please provide a whiteboard on inverse trigonometric functions within implicit differentiation. Looked around the internet and couldn't find anything specific. Thank you.
k2L93 1 year ago
Having a problem with the second and final portion of this problem:
Find the function g(x) that is implicitly defined by the relation:
sqrt x^2 - y^2 + Arc cos x/y = 0 , y dosen't equal 0,
Answer: y = g(x) = x.
Someone posted the following for the answer:
g'(x)= 1/2 (x^2-y^2)^-1/2 ( 2x-2ydy/dx) + -1/sqrt(1-(x/y)^2) [ 1/y][ -x/y^2dy/dx]=0
Was able to work up to and complete the first part using the chain rule. The answerer through me off with the cos inverse function with the y^2.
k2L93 1 year ago