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

In summary, the problem asks to show that if f is a differentiable function such that f(g(x)) = x and f'(x) = 1 + [f(x)]^2, then g'(x) = 1/(1+ x^2). The solution involves using differentials and variables instead of functions, and showing that the derivative of the inverse function can be found by taking the reciprocal of a ratio of differentials. This may be a challenging question for first year calculus students.
  • #1
sporus
21
0

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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  • #2
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
 
  • #3
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?
 
  • #4
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).
 
  • #5
LoopQG said:
f'(x)=1+[f(x)]^2

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

This is incorrect.
 
  • #6
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?
 

1. What is difficult differentiation?

Difficult differentiation is a concept in mathematics where a function is differentiated multiple times. It often involves complex or abstract functions that can be challenging to differentiate using traditional methods.

2. Why is differentiation considered difficult?

Differentiating a function can be difficult because it requires a deep understanding of mathematical concepts and techniques. It also involves multiple steps and can become increasingly complex with each differentiation.

3. How do you approach a difficult differentiation question?

To approach a difficult differentiation question, it is important to first understand the given function and its properties. Then, you can use various techniques such as the chain rule, product rule, or quotient rule to differentiate the function step by step.

4. What are some common mistakes when solving difficult differentiation questions?

Some common mistakes when solving difficult differentiation questions include incorrect use of differentiation rules, forgetting to apply the chain rule, and miscalculating derivatives. It is important to carefully check each step and practice regularly to avoid these mistakes.

5. How can I improve my skills in solving difficult differentiation questions?

To improve your skills in solving difficult differentiation questions, it is important to practice regularly and familiarize yourself with various differentiation techniques. You can also seek help from textbooks, online resources, or a tutor to better understand the concepts behind these questions.

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