Difficult differentiation question (the concept behind this question is elusive)

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Homework Help Overview

The problem involves a differentiable function f and its relationship with another function g, where f(g(x)) = x and f'(x) = 1 + [f(x)]^2. The goal is to demonstrate that g'(x) = 1/(1 + x^2). The context is centered around differentiation and the properties of inverse functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between f and g, with some suggesting that f is the inverse of g. There are attempts to express g'(x) in terms of f and its derivative, and some participants raise concerns about the use of integration in their approaches. Others explore the use of differentials and Leibniz's notation to clarify the relationship between the derivatives.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have offered insights into using differentials, while others express uncertainty about the appropriateness of certain methods given their current curriculum. There is no explicit consensus on a single approach, but multiple lines of reasoning are being examined.

Contextual Notes

One participant notes that they have not yet learned integration, which may limit their ability to approach the problem in certain ways. There is also a question raised about the fairness of the problem for first-year calculus students.

sporus
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Homework Statement



suppose f is a differentiable function such that f(g(x)) = x and f'(x) = 1 + [f(x)]^2. Show that g'(x) = 1/(1+ x^2)

Homework Equations





The Attempt at a Solution



since f(g(x)) = x, i think that f is the inverse of g.
so f = g{inverse}

f'(x) = f'(g(x)) * g'(x) = 1 + [f(x)]^2

we are given g'(x) and f'(x), but i can't make the connection between the fact that f = g{inverse} and how it affects f'(g(x)) because that is the only missing part of the equation.
 
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f'(x)=1+[f(x)]^2

so f(x)=x+[f(x)^3]/3


so take f'(g(x))

following the rule stated

f'(g(x))=1 + x^2 = 1/g'(x)

notice also f'(g(x))=1 so g(x) must subtract out that x^2

hope this helps
 
we haven't learned any integration yet. we only do derivatives in calc 1 so i might not get marks for solving using integration. is there a way to do this without using integration?
 
sporus said:
suppose f is a differentiable function such that f(g(x)) = x and f'(x) = 1 + [f(x)]^2. Show that g'(x) = 1/(1+ x^2)

Here is how I would tackle it work with differentials and use variables instead of functions.

Let y = g(x) so f(y)=x.

g'(x) = dy/dx then f'(y) = dx/dy

Then rewrite: f'(y)=1+f(y)^2 = 1+x^2.
dx/dy = 1+x^2
dy/dx = 1/(1+x^2)

Differentials can be defines so that Leibniz's notation is perfectly consistent even up to taking the reciprocal of a ratio of differentials (which gives the derivative of the inverse function).
 
LoopQG said:
f'(x)=1+[f(x)]^2

so f(x)=x+[f(x)^3]/3

This is incorrect.
 
jambaugh said:
Here is how I would tackle it work with differentials and use variables instead of functions.

Let y = g(x) so f(y)=x.

g'(x) = dy/dx then f'(y) = dx/dy

Then rewrite: f'(y)=1+f(y)^2 = 1+x^2.
dx/dy = 1+x^2
dy/dx = 1/(1+x^2)

Differentials can be defines so that Leibniz's notation is perfectly consistent even up to taking the reciprocal of a ratio of differentials (which gives the derivative of the inverse function).

i'll try to wrap my head around this one. thanks
btw, do you think this is a fair question for first year calc?
 

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