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$$f(x) = \sin(x) + 2x + 1$$

After setting

$$x = \sin(y) + 2y +1$$

I don't know how to solve for $y$.

For context, the problem is asking to find $g'(x)$, knowing that $g(x)$ is the inverse of $f(x)$.

I thought maybe I could differentiate and find the inverse of $f'(x)$. Would that be $g'(x)$?

Thanks!

Gary
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Richard
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  • Various solutions for inverse here or use derivative of inverse function – Тyma Gaidash Oct 16 '23 at 02:12
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    I did the math formatting for you. Enclose equation lines in double dollars $$...$$ and inline equations in single dollars $...$. Use built in functions like \sin and \cos for proper formatting, and do fractions with \frac{a}{b}. In general curly braces are used for grouping { .. }, for example, to make a subscript use x_{i+1} that renders like $x_{i+1}$. – John Alexiou Oct 16 '23 at 02:24
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    Do you know the inverse function theorem?

    https://en.wikipedia.org/wiki/Inverse_function_theorem

     – user52817 Oct 16 '23 at 02:25
  • First of there isn't an analytical function for the inverse of functions of this form $\sin(x) + a x$.

    What you are probably asked to do is implicit differentiation.

     – John Alexiou Oct 16 '23 at 02:30
  • Thank you, @JohnAlexiou! Indeed, I believe this is an implicit differentiation problem. I'm just struggling to make the connection. If I differentiate implicitly and isolate y', what do I have in relation to g(x), aka the inverse? – Richard Oct 16 '23 at 02:45

2 Answers2

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The point of this exercise is the Inverse Function Theorem, which allows you to find the derivative of the inverse function at particular points under certain conditions (the link noted above in the comments explains this exactly so I will not state all that). It is particularly useful when you cannot find the inverse $g(x)$ explicitly using algebra as is the case here, and furthermore you should not generally expect to obtain $g'(x)$ as an explicit function.

A simpler example where we can find the inverse explicitly will help illustrate. Consider $f(x)=\frac 1{x-1}$. Then the inverse $g(x)$ is given by $g(x)=\frac 1x +1$. Obviously we have to exclude $x=1$ but this is fine: the IVT says we can operate in a small interval around any $x \ne 1$.

Since $g(f(x)) = x$, the chain rule gives that:

$g'(f(x))f'(x) =1$ or $g'(f(x)) = \frac 1{f'(x)}$.

You can already see that this generally will not allow you to find $g'(x)$ as an explicit function; however this formula does hold at individual points.

Back in the example, we have $f(2)=1$ and $g(1)=2$.

This tells us that $g'(1)$ should be $1/f'(2)$. Since $f'(x) = \frac {-1}{(x-1)^{2}}$, $f'(2)=-1$.

In this case we can check: $g'(x) = -\frac 1{x^2}$ so $g'(1)=-1$.

Note that that $g'(x) \ne \frac 1{f'(x)}$ as functions which is a common misconception.

You should be able to apply this in your example now to get a formula for $g'(f(x))$.

AlgTop1854
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Inverse functions are found by doing all of the steps that were done to the original function in reverse order. For example, to solve 2x + 3 = 11, think about how the function was constructed using PEMDAS. X was multiplied by 2 (using a "getting dressed" analogy, "put on your socks") then 3 was added ("put on your shoes"). To solve it, subtract 3 ("take off your shoes") then divide by 2 ("take off your socks"). So if in the original function, the instructions are "take the sine, then add 2x, then add 1", to find the inverse, subtract 1, then subract 2x, then take the inverse sine/arcsine of this answer. So it should be $f^{-1}(x) = \arcsin(x-1-2x) = \arcsin(-x-1)$. I think.

Nate
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    This is incorrect since clearly $f(\arcsin(-x-1)) \neq x$. – Gary Oct 16 '23 at 02:48
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    This isn't right. "Add $2x$" isn't a simple operation once you've already applied $sin$ to to the original $x$, so you're not just applying these step by step. The entire function is more complicated and has an inverse that is much harder to describe. You can check by seeing that applying this function to original does not get you back where you started. – Chessanator Oct 16 '23 at 02:50
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    Originally $f(x) = \sin(x) + 2x + 1$, "subtract 1, then subract 2x, then take the inverse sine/arcsine" gives $$\arcsin (f(x)-1-2x) = x$$ or $$f^{-1}(x) = \arcsin(x-1-2f^{-1}(x))$$. How do you plan to remove the $x$ on the top left and the $f^{-1}(x)$ on the bottom right? – peterwhy Oct 16 '23 at 02:50
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    Your proposed $f^{-1}(x) = \arcsin(-x-1)$ is undefined for large $x$. – peterwhy Oct 16 '23 at 02:57
  • @Nate Have you considered deleting this "answer"? – Gary Oct 21 '23 at 12:17