Derivative of Inverse Function

[sydneyfranke]

Homework Statement

Find g' (-1/2), where g(x) is the inverse of f(x) = x³ / (x² + 1)

Homework Equations

g'(x) = 1/f' [g(x)]

The Attempt at a Solution

Everything else I tried got really messy, but I feel like I'm going out on a limb by:

1) y=x³ / (x²+1)

2) y=x³x^-2 / 1

3) y=x

4) g '(0.5)= 1/ (0.5³ /(0.5²+1) = 10

I don't even know if I'm going about it the right way. Any help would be awesome. Thanks.

 

[Inferior89]

You say in 1) that
y = x^3 / (x^2 + 1)
and in 3) that
y = x.

If you put in x = 1 you see that in 1) you get 1/2 and in 3) you get 1. Something must hence be wrong.

 

[sydneyfranke]

Very true. I realize something is wrong. That was just my attempt.

 

[sydneyfranke]

But . . . I could still use some help with this problem, obviously.

 

[fzero]

1) y=x³ / (x²+1)

2) y=x³x^-2 / 1

Line 2 is incorrect, it should be

y=x³x^-2 / (1+x^-2),

which actually just makes things more complicated. Just differentiate the expression on line 1 directly.

 

[JonF]

It would help to figure out what g(-1/2) was equal to

Then you could just plug stuff into this relationship: g'(x) = 1/f' [g(x)] -> g'(-1/2) = 1/f' [g(-1/2)]

Since we know that g is the inverse of f, we should try to solve when f(a) = -1/2. Then we would know a = g(-1/2).

So -1/2 = x^3 / (x^2 + 1) -> -1/2*(x^2 + 1) = x^3 -> 0 = x^3 + 1/2x^2 + ½ which clearly has a solution at -1. So we know g(-1/2) = -1 which means: g'(-1/2) = 1/f'(-1) and you can do the rest from there id bet.

 

[sydneyfranke]

So I differentiate to get

(x²+1(3x²))-(2x(x³))/((x²+1)²)

And then I dared to simplify it down to

(x⁴+3x²)/((x²+1)²)

But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.

 

[Inferior89]

As the guy above said, if f(c) = -1/2 then g(-1/2) = c. To find out what g(-1/2) is we need to find c, that is to solve the equation f(x) = -1/2.

When this is done you can just plug everything into the formula.

 

[jackmell]

So I differentiate to get

(x²+1(3x²))-(2x(x³))/((x²+1)²)

And then I dared to simplify it down to

(x⁴+3x²)/((x²+1)²)

But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.

Surely if I have a function y=f(x), then the inverse is just x=g(y) and it's derivative is \frac{dx}{dy} which is a function of y. So that for:

f(x)=\frac{x^3}{x^2+1}

we have x=f^{-1}(y) and so the derivative of the inverse at y_0 can be written as:

\frac{df^{-1}}{dy}\biggr|_{y_0}=\frac{dx}{dy}\biggr|_{y_0}=\frac{1}{\frac{dy}{dx}}\biggr|_{x(y_0)}

So now let y_0=-1/2[/tex]. Then solve:<br /> <br /> y_0=-1/2=\frac{x^3}{x^2+1} <br /> <br /> which has three solutions. The inverse and it&#039;s derivative are multi-valued right since it&#039;s going to be a cube root correct? Therefore, just find the three roots (x_1,x_2,x_3) and for example, then one derivative of the inverse at y=-1/2 is:<br /> <br /> \frac{df^{-1}}{dy}\biggr|_{y=-1/2}=\frac{dx}{dy}\biggr|_{y=-1/2}=\frac{1}{\frac{dy}{dx}}\biggr|_{x(-1/2)=x_1}<br /> <br /> Try and follow the following Mathematica code which explicitly computes the inverse, then finds it&#039;s (multivalued) derivative at y=-1/2:<br /> <br /> <div class="bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code"> <div class="bbCodeBlock-title"> <i class="fa--xf fal fa-code "><svg xmlns="http://www.w3.org/2000/svg" role="img" aria-hidden="true" ><use href="/data/local/icons/light.svg?v=1756418028#code"></use></svg></i> Code: </div> <div class="bbCodeBlock-content" dir="ltr"> <pre class="bbCodeCode" dir="ltr" data-xf-init="code-block" data-lang=""><code>f[x_] := x^3/(x^2 + 1); mysols = x /. Solve[f[x] == y, x] N[D[mysols, y] /. y -&gt; -2^(-1)]</code></pre> </div> </div>