Hi all. I am working on the derivative of inverse trig functions. I would like to know how important is it that I understand the proof of the derivative of sine and cosine inverse. I have the proof in front of me but I do not understand it AT ALL. I am nervous about moving on with the lesson because I think I would need to understand the concept. Please advise. Thanks!
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Proof of derivative of sine and cosine inverse
- Thread starter mikagurl
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Hi all. I am working on the derivative of inverse trig functions. I would like to know how important is it that I understand the proof of the derivative of sine and cosine inverse. I have the proof in front of me but I do not understand it AT ALL. I am nervous about moving on with the lesson because I think I would need to understand the concept. Please advise. Thanks!
y = sin-1(x)
x = sin(y) ...................................(1)
\(\displaystyle \frac{dx}{dy} \ = \ \) cos(y)
\(\displaystyle \frac{dx}{dy} \ = \ \sqrt{1 - x^2}\) ............from (1) using sin2(y) + cos2(y) = 1
\(\displaystyle \displaystyle{\frac{dy}{dx} \ = \ \frac{1}{\sqrt{1 - x^2}}}\)
If you are still stuck - let us know where.
Yes, you can use the same theory with the other trig functions. Another way which, of course, leads to the same result:Thanks. I understand it now!!! So I can use the same theory with cosine as well??? I'm guessing.
y=sin-1(x)
x = sin(y)
dx/dx = 1 = d(sin(y))/dy dy/dx = cos(y) dy/dx
by the chain rule or
dy/dx = 1/cos(y) = 1 / \(\displaystyle \sqrt{1 - x^2}\)