Calculate second derivative of a function in terms of another function

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SUMMARY

This discussion focuses on calculating the second derivative of an inverse function, specifically ##g''(x)## in terms of the original function ##f## and its derivatives. The user successfully derived the first derivative ##g'(y)## and attempted to compute ##g''(y)## but struggled to eliminate terms involving ##g''(f(x))## and ##g'(f(x))##. For part b, the user is tasked with proving that if ##f'(x) > 0## and ##f''(x) > 0##, then ##g''(x) < 0##, which relates to the properties of inverse functions and their behavior under monotonicity.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and inverse functions.
  • Familiarity with the concepts of bijective functions and class ##C^2## functions.
  • Knowledge of the chain rule and its application in differentiation.
  • Basic principles of monotonicity and its implications for inverse functions.
NEXT STEPS
  • Study the properties of inverse functions, particularly how derivatives relate between a function and its inverse.
  • Learn about the implications of monotonicity on the behavior of functions and their inverses.
  • Explore geometric interpretations of inverse functions, especially reflections across the line y = x.
  • Investigate examples of functions with known inverses, such as quadratic functions, to solidify understanding of their behavior.
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Students and educators in calculus, mathematicians interested in differential calculus, and anyone studying the properties of inverse functions and their derivatives.

mahler1
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Homework Statement .

Let ##f:[a,b] \to [\alpha,\beta]## be a bijective function of class ##C^2## with an inverse function also of class ##C^2## ##g:[\alpha,\beta] \to [a,b]##.

a)Calculate ##g''(x)## for every ##x \in (\alpha,\beta)## in terms of ##f## and its derivatives.
b)If ##f'(x)>0## and ##f''(x)>0## for every ##x \in (a,b)##, prove that ##g''(x)<0## for every ##x \in (\alpha,\beta)##

For part ##a)##, all I could do was:

##g(y)=(g\circ f)(x)##, so ##g'(y)=(g\circ f)'(x)=g'(f(x))f'(x)##.
Then, ##g''(y)=\dfrac{d}{dx}g'(y)=\dfrac{d}{dx}g'(f(x))f'(x)##.

Computing this derivative gives:
##\dfrac{d}{dx}g'(f(x)).f'(x)=g''(f(x){f'(x)}^2+g'(f(x))f''(x)##

Now, I need to find ##g''(y)## just in terms of ##f## and its derivatives, I don't know how to get rid of ##g''(f(x))## and ##g'(f(x))##.

For part ##b)## I have no idea what to do, I would appreciate any suggestion or idea to prove ##b)##.
 
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What you are missing is that g(f(x)) = x. Plug that into your first equation, and you will find things simplify out nicely.

What you computed so far is just d/dx(f(g(x)) without using the fact that these are inverses.

Part b is about the nature of inverses. For example, if f is continuous, so is its inverse; if discontinuous so is its inverse, if differentiable once, so is its inverse, etc.

Here's how to do some thinking about things:

What if f is monotonic increasing? What could you say about f inverse? Take some examples -- look at f(x) = ##x^2## on [0,1]. What is its inverse there? Does f increase or decrease ? What about its inverse? Based on this one example would you feel safe generalizing? Or would you feel safe after looking at 2 or 3 examples? I realize anything can happen on example 4, but when you look at these examples, you can probably see a reason why things are working out as they do; and that reason would apply to all invertible functions.

Once you get that far, look at your computation for g'(x) in terms of f. Does that bear out the intuitive conclusions you drew above?

Okay, go on to the f'' case. In the example above what happens to g''? Why?

Now you can construct an argument about f'' > 0 implying g'' < 0. You might use your computation in part a. Or you might imagine a geometric argument about f inverse being the reflection of f across the line y = x.
 
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