Why do I have to use both quotient and chain rule for this problem? | Mathematics | Page 1

golden roverBOT Oct 19, 2023, 3:35 PM

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left elbow Oct 19, 2023, 4:37 PM

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y(x) is a function of x

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y(x) = ln(x) + e^x

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So if you compute the derivative of y wrt x, you would write dy/dx

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In this problem they are not asking you for the derivative of y

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They are asking you for the derivative of
$y^{-1}(x)$

pearl knotBOT Oct 19, 2023, 4:39 PM

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Daddy_314

left elbow Oct 19, 2023, 4:39 PM

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Which is the inverse function of y

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So if for example we had

$y = x^2$ for x positive

$y^{-1}(x) = \sqrt{x}$ is the inverse function of $y$

pearl knotBOT Oct 19, 2023, 4:40 PM

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Daddy_314

left elbow Oct 19, 2023, 4:40 PM

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Now the reason you are lost

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Is because they didnt call this inverse function $y^{-1}(x)$ but they called it $x(y)$

pearl knotBOT Oct 19, 2023, 4:41 PM

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Daddy_314

left elbow Oct 19, 2023, 4:42 PM

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The reason they do this is because if
$y = x^2$ for example, then
$x = \sqrt{y}$ for x positive,

This means we can express x as a function of y but this is just a way to trick you and make you sad

pearl knotBOT Oct 19, 2023, 4:42 PM

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Daddy_314

left elbow Oct 19, 2023, 4:42 PM

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No one cares if you call it $x(y)$ or $y^{-1}(x)$

pearl knotBOT Oct 19, 2023, 4:42 PM

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Daddy_314

strong flame Oct 19, 2023, 4:47 PM

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left elbow No one cares if you call it $x(y)$ or $y^{-1}(x)$

Ah I see, but like I still don't really get why I need to use the chain rule for that case?

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Or is it that if we were to find the derivative of y i'd only need to use the quotient rule?

left elbow Oct 19, 2023, 4:47 PM

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Yes

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We get there now

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Heres the magical trick:

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When we want to find inverse function derivatives

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We use the chain rule

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Let me give you an example

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If $y(x) = x^2$
And $g(x) = y^{-1}(x) = \sqrt{x}$
Imagine we want to find the derivative of g

pearl knotBOT Oct 19, 2023, 4:49 PM

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Daddy_314

left elbow Oct 19, 2023, 4:49 PM

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We use the fact that by definition,

$f(g(x)) = x$ for all x

pearl knotBOT Oct 19, 2023, 4:49 PM

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Daddy_314

left elbow Oct 19, 2023, 4:50 PM

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Therefore differentiating both sides of this equality

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$g'(x) \times f'(g(x)) = 1$

pearl knotBOT Oct 19, 2023, 4:53 PM

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Daddy_314

left elbow Oct 19, 2023, 4:53 PM

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Since f'(x) = 2x for all x

Then
$g'(x) \times 2 \sqrt{x} = 1$
And $g'(x) = \frac{1}{2 \sqrt{x}}$

pearl knotBOT Oct 19, 2023, 4:53 PM

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Daddy_314

left elbow Oct 19, 2023, 4:54 PM

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We have found the derivative of the inverse function !

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Their notation is truly horrible

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Because they wrote:

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dy/dx = 1/(dx/dy)

left elbow Oct 19, 2023, 4:55 PM

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left elbow $g'(x) \times f'(g(x)) = 1$

When in reality, they mean this

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This is why there's a dx/dy = 1/(dy/dx))

strong flame Oct 19, 2023, 4:59 PM

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But this seems so much more complex than just taking the derivative of y^-1 = sqrt(x)

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Although, it may not always be possible to find the inverse so easily

left elbow Oct 19, 2023, 5:01 PM

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Well

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This was just an illustration

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Both square root and x squared have easy derivatives

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But sometimes things are much harder

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This method is so powerful it allows you to compute derivatives of

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$arccos(x)$

pearl knotBOT Oct 19, 2023, 5:02 PM

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Daddy_314

left elbow Oct 19, 2023, 5:02 PM

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Or any other function

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Whose derivative is known

strong flame Oct 19, 2023, 5:04 PM

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Yeah, I see. This is just the inverse function theorem right? I am still trying to wrap my head around it, because this example is quite easy I am not too sure if I really understand it or just this example

strong flame Oct 19, 2023, 5:05 PM

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left elbow This method is so powerful it allows you to compute derivatives of

But in general, how should I proceed to solve derivatives of inverse functions? Because, as you said, it is not always easy to find the inverse

left elbow Oct 19, 2023, 5:11 PM

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The inverse function theorem

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Is based on the chain rule

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Exactly

strong flame Oct 19, 2023, 5:14 PM

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Yeah, so if I have a function f(x), and I want to find the derivative of its inverse. Should I use 1/f'(x) (i.e. 1/(dy/dx))?

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Because, that makses sense, but after that I'm stuck

left elbow Oct 19, 2023, 5:16 PM

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Be careful,

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And this is exactly the problem

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With their notation

strong flame Oct 19, 2023, 5:48 PM

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left elbow Be careful,

Could you elaborate a bit on that?

left elbow Oct 19, 2023, 5:50 PM

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I think they are being

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Dishonest

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In your solution

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They are expressing the derivative of the inverse function not in terms of the independent variable, but in terms of the inverse function itself

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Here's how:

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Let
f(x) = ln(x) + e^x
And g(x) its inverse

g'(x) * (f'(g(x)) = 1

With:
f' = 1/x + e^x
f'(g(x)) = 1/g + e^g

g'(x) * (1/g + e^g) = 1

g'(x) = 1/(1/g(x) + e^g(x))

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This means the derivative of g has been expressed implicitly using the function g

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This is what they wrote at first

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When they wrote
Dx/dy = ... = x/(1+x e^x)

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The x is not a variable, its a function !

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g' is dx/dy

and g is x(y)

left elbow Oct 19, 2023, 5:54 PM

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strong flame But in general, how should I proceed to solve derivatives of inverse functions? ...

Which makes your remark here very nice

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As the inverse of f cannot be explicitly expressed

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We work with it "implicitly"

strong flame Oct 19, 2023, 5:56 PM

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Hmm, that's very strange. I doubt my professor would do something like this on purpose.

left elbow Oct 19, 2023, 5:57 PM

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Its not on purpose

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Nor not on purpose

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Its something that isnt clear in their solution

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And that I am making clear for you

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x is not a variable

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Its a function in what they wrote

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This means dx/dy = x/(1+x e^x)

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Is not what you think it is

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Because x is a function

strong flame Oct 19, 2023, 5:58 PM

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Is g'(x) * f'(g(x)) = 1, something I should always be doing for the inverse function theorem though? Because then I get g'(x) = 1/f'(g(x)), which is what the definition of the inverse function theorem states as well

left elbow Oct 19, 2023, 5:59 PM

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left elbow Let f(x) = ln(x) + e^x And g(x) its inverse g'(x) * (f'(g(x)) = 1 With: f' = 1...

This is the real meaning behind their notation

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g' was expressed in terms of g(x)

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This was not the case for f' which we did not express in terms of f

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But in terms of x

strong flame Oct 19, 2023, 6:00 PM

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Okay, but just in general, when it comes to finding the derivative of an inverse function, how should I tackle that?

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Because now i'm feeling more cluelesss than before lol

left elbow Oct 19, 2023, 6:01 PM

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Its normal to feel clueless after learning the truth

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They were trying to make it seem like it was something else

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I dont like the dy/dx and dx/dy notation

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Because it doesnt mean what you think it means

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Just to clarify

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Again

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They want you to express the second derivative of the function x(y)
In terms of the function x(y)

Not in terms of y

strong flame Oct 19, 2023, 6:04 PM

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Yeah, but I am not talking about the exercise anymore. Just in general, if I need to determine the derivative of the inverse function, how should I do it? Because now I am not too sure I understand how to tackle a problem like that

left elbow Oct 19, 2023, 6:11 PM

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I only use the chain rule in the form I showed you

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And I have a fulfilling life

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Pick one form of the theorem that you can master

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Namely,
(f(g(x))' = g'(x) f'(g(x))

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This is the one formula to learn

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All the other ones you can put in the garbage

strong flame Oct 19, 2023, 6:15 PM

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Okay cool, thanks for that

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Is f(x) the normal function and g(x) its inverse?

left elbow Oct 19, 2023, 6:16 PM

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It probably doesnt matter

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You can choose f and g

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If g is inverse of f
f is inversr of g

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The only problems are domains of definition

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But

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In practice, you can choose f and g as you want

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If you wanna find the derivative of the inverse function
Its a bit more clever to use f(g(x))
And take the derivative

strong flame Oct 19, 2023, 6:18 PM

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But if I get a function f(x), I need to find its inverse (g(x)), and then I can make the composition and differentiate it?

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To get the derivative of the inverse?

left elbow Oct 19, 2023, 6:23 PM

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Yes

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And if you cant explicitly find the inverse

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You fall into a similar case

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As this exercise

strong flame Oct 19, 2023, 6:25 PM

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Then I just take derivative of f(x), 1/(dy/dx) and use both chain and quotient rule?

left elbow Oct 19, 2023, 6:26 PM

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Please dont use dy/dx

strong flame Oct 19, 2023, 6:27 PM

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Oops, 1/f'(x) then

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Why not 1/(dy/dx) though? Or is that too complicated to explain now?

strong flame Oct 19, 2023, 7:23 PM

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left elbow Please dont use dy/dx

Could you maybe help me with a different problem as well?

left elbow Oct 19, 2023, 7:24 PM

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Yes

strong flame Oct 19, 2023, 7:46 PM

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left elbow Yes

Cool. How do I tackle a problem like this?

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Like, I know it is bounded, so it must eventually be really small. But I am not really sure how to express it

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And I need the first one for the others

left elbow Oct 19, 2023, 7:51 PM

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f is continuous so indeed it is bounded on the compact [0;1]

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We may then write
|f(x)| <= M for all x in [0;1]

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You could write it equivalently like:
$ a \leq f(x) \leq b$ for all x in $[0;1]$
This reveals that for all x in [0;1]

$a x^n \leq x^n f(x) \leq b x^n$

pearl knotBOT Oct 19, 2023, 7:54 PM

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Daddy_314

left elbow Oct 19, 2023, 7:55 PM

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What happens when you integrate those three functions

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The leftmost and rightmost functions can be easily integrated

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And then you can apply the squeeze theorem

strong flame Oct 19, 2023, 8:19 PM

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left elbow The leftmost and rightmost functions can be easily integrated

Just antiderivative power rule right?

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but wouldn't that mean it is 1? The answer is supposed to be 0

left elbow Oct 19, 2023, 9:08 PM

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What is 1?

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What's $[\frac{x^{n+1}}{n+1}]^1_0$

pearl knotBOT Oct 19, 2023, 9:10 PM

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Daddy_314

left elbow Oct 19, 2023, 9:10 PM

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Definitely not 1, the result depends on n

strong flame Oct 19, 2023, 9:48 PM

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left elbow Definitely not 1, the result depends on n

Wait, we're evaluating it at x = 1 and x = 0 correct?

strong flame Oct 19, 2023, 9:49 PM

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left elbow What's $[\frac{x^{n+1}}{n+1}]^1_0$

As x is between 0 and 1, the denominator will get really big regardless (n -> infinity) so it will end up 0?

left elbow Oct 19, 2023, 11:26 PM

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You didnt evaluate

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The lower and upper bounds

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Of the inequality

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Before taking the limit

strong flame Oct 20, 2023, 7:27 AM

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left elbow Before taking the limit

Won't it just end up being 1^(n+1)/(n+1) - 0^(n+1) / (n+1) = 1^(n+1)/(n+1) = 1/(n+1)?

left elbow Oct 20, 2023, 8:35 AM

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Yes

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So a/(n+1) <= integral(x^n f) on 0,1 <= b/(n+1)

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And taking the limit here as n goes to infinity

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U get

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By the squeeze theorem

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(= the limit sandwich theorem)

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That the middle term goes to zero

strong flame Oct 20, 2023, 8:40 AM

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Ah like that, I see.

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It's more clear now, thanks

left elbow Oct 20, 2023, 8:51 AM

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We used the fact that

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The integral from a to b

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Preserves inequalities

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(This is why we integrated the three functions)

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From 0 to 1

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P(x) < f < Q(x)
Implies integral P < integral f < integral Q

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We just took P(x) = ax^n

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And q = b x^n

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So u can mention this

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To get all the points

#Why do I have to use both quotient and chain rule for this problem?