Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm watching this video lecture http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-11-chain-rule/ and I'm stuck at around 3:40, I can't seem to figure out what he is doing.

He is showing how to derive $f(x)=\sin^{-1}(x)$.

At some point he goes from the expression $\frac{dy}{dx}=\frac{1}{\cos(y)}$ to $\frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}$.

Ok, I know that's the result, that's how I always did it, but I never actually derived it myself. So yeah, I'd like to know what he did in those last two steps, to get from the first expression to the second.

I know it may (will) be something completely stupid and I'll say 'oh... facepalm', but for some reason I can't figure out how he did it. I'm guessing the next logical step is to replace $y=\sin^{-1}(x)$, but then? I've never been in good terms with trigonometric functions and identities really, so I'd appreciate some enlightening.

Thank you.

share|improve this question
add comment

2 Answers 2

Remember that (in a suitable interval) $\cos y = \sqrt{1-\sin^2 y}$, and that here $y = \arcsin x$, so $x = \sin y$ and $\cos y = \sqrt{1-x^2}$. $\frac{dy}{dx} = \frac1{\cos y}$ comes from the "formula" $\frac1{\frac{dx}{dy}} = \frac{dy}{dx}$.

share|improve this answer
add comment

Another way to deal with this, which is useful for obtaining all of the derivatives of the inverse trig functions, is to keep in mind that $\ y \ = \ \sin^{-1} x \ \Rightarrow \sin y \ = \ x \ = \frac{x}{1} \ $. You can now construct a right triangle with one of the angles being $y$ : the leg opposite $y$ has length $x$, and the hypotenuse, a length of $1$ . The leg adjacent to $y$ must then have length $\sqrt{1 - x^2}$ . Hence,

$$\cos y \ = \ \frac{\sqrt{1 - x^2}}{1} \ \Rightarrow \frac{d}{dx}\sin^{-1} x \ = \ \frac{1}{\sqrt{1 - x^2}} .$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.