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Differentiate: G(x)=sqrt(1-x^2) * arccos(x)
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I got \[G'(x)=-\frac{ \arccos(x) }{ \sqrt{1-x^2} }-1\] but the book says there should be an extra x in the numerator
know about the chain rule ?
Yes, I'm guessing that's where I slipped
d/dx (sqrt(1-x^2)) = 1/2 sqrt (1-x^2) times d/dx (1-x^2) and whats d/dx (1-x^2) = ...?
I thought d/dx(1-x^2) is just 2 no?
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I mean -2
d/dx x^2 = ... ?
*facepalm* 2x
I was thinking the x turns into a 1 and leaves 2 alone, thanks
welcome ^_^
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