I'm asked to find the derivative of $\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$ using implicit differentiation. The question is why do I need to use implicit differentiation when I have a function defined only in terms of $x$.
Asked
Active
Viewed 61 times
0
-
Who says implicit differentiation plays here? This is a straghtforward application of the theorem about the derivative of an inverse function... – DonAntonio May 21 '17 at 16:38
-
My assignment the teacher asked us to solve the problem using implicit differentiation, that's what confuses me – Andres Romero May 21 '17 at 16:40
-
Well if you let $y = \arcsin x$, then $x = \sin y$. Now differentiate with respect to $x$. What would you get? – Alex Vong May 21 '17 at 16:55
2 Answers
1
We use implicit differentiation when we don't have $y$ given as an explicit function of $x$, i.e. we do not have $y=\mathrm f(x)$.
The trick to find the derivative of $\arcsin x$ is to write $y=\arcsin x$, but then re-write as $x=\sin y$. In this case, we do not have $y$ as an explicit function of $x$; it is given implicitly.
You really think of $y$ as being a function of $x$, say $y(x)$, and then apply the chain rule to both sides of the equation: $$x=\sin [y(x)] \ \ \ \implies \ \ \ 1 = \cos [y(x)]\cdot\frac{dy}{dx}$$
This gives $\frac{dy}{dx} = \frac{1}{\cos y}$ and now you need to write $\cos y$ as a function of $x$, given that $x=\sin y$.
Note: For $\arcsin$ to be well-defined, you need to specify the domain of $\sin$. You need a function to be one-to-one for it to have an inverse. If you have restricted sine to $-\frac{1}{2}\pi \le x \le \frac{1}{2}\pi$ then the derivative of its inverse is $(1-x^2)^{-1/2}$. But if you restrict sine to $\frac{1}{2}\pi \le x \le \frac{3}{2}\pi$ then the inverse has derivative $-(1-x^2)^{-1/2}$.
Fly by Night
- 32,272
1
Let, $x=sin\theta \implies \theta=sin^{-1}x.$
$\frac{dx}{d\theta}=cos\theta\implies\frac{d\theta}{dx}=\frac{1}{cos\theta}\implies\frac{d}{dx}(sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
Dharmaraj Deka
- 327