Have You Considered This Trick for Finding the Derivative of a Cubic Function?

[Robokapp]

I'm guessing not a lot will care about this becasue it's not very relevant, but my calc teacher couldn't do this and I did it in a few seconds, so i'll expose it.

The problem stated that f(x)=x^3+x

and inverse of f(x)=g(x) and g(2)=1

question: Find g'(2)

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My teacher tried to create a formula to connect inverse derivative answers and inverse functions for cubics. He couldn't. So while staring at it I realize how the derivative is dy/dx which is appearing everywhere you derivate a y.

so I write y=x^3+x
take inverse x=Y^3+y
and I don't care about what the function looks like. I don't worry about putting it in standard form like he tried. I keep it like this and take derivative. Out of nowhere I might say I had written down 1=3y^2 dy/dx + dy/dx

and isolating the dy/dx => dy/dx = 1/(1+2y^2)

since point (2,1) was given, the fact that I have no x is not important. i can plug in y instead. And I get the final answer. g'(2)=1/4

The relevance of this is that finding the derivative of a function can be expressed in many forms, related to various letters in that expression. many times the y' has both x and y.

But...Standard form was not important here, and pretty much everyone, myslef included for a few minutes were hooked up on putting it in standard form...

I thought i'd share this with you.

~Robokapp

 

[DaMastaofFisix]

Very impressive, I wish I had that sort of intuitive math skill. See If you can extend that particular problem to a more genralized formula or set of steps. If you can find some overarching shortcut, you could save yourself a whole lot of time. Try additional polynomial terms and there inverses. Tell us what happens
-see...we care-

 

[philosophking]

but i don't think i understand. how is dy/dx [ y^3 ] = 3y^2? isn't it zero?

 

[masudr]

so I write y=x^3+x
take inverse x=Y^3+y

Substituting x into the first equation gives:

(y^3+y)^3 + y^3+y\ne y[/itex]<br /> <br /> So how is that the inverse?

 

[Hurkyl]

Given the relation y=x³+x, which defines the graph of y=f(x), its inverse relation is x=y³+y, which defines the graph of y=g(x).

Or, to write it more clearly: (IMHO)

We have that f(x)=x³+x (for all x in the domain of f), so we also have that x=g(x)³+g(x). (for all x in the domain of g)


You've discovered and applied one of the powerful methods of differential calculus. Congratulations!


Now, the next question is to generalize.

For any function f, whose inverse is g, can you write down a formula for g'(a)?

 

[masudr]

Is that what I would perhaps call implicit differentiation?

 

[Hurkyl]

Yep! (Well, one of the major steps in this calculation is called implicit differentiation)

 

[Robokapp]

but i don't think i understand. how is dy/dx [ y^3 ] = 3y^2? isn't it zero?

well, y is a variable, not a constant.

let's pick y=2x => dy/dx=2 agreed?

but it is that because dy/dx is the derivative of y. basically you derivate both sides. Everywhere you see a y you add a dy/dx at the end of it because it doesn't contain an x. y' or dy/dx is same thing...

 

[masudr]

Yep! (Well, one of the major steps in this calculation is called implicit differentiation)

Silly teacher didn't know implicit differentiation! What level is the original poster at? Congratulations to him/her for working this out for him/herself, though.

 

[Robokapp]

hah. I got it!

I'm Calculus AB Highschool senior by the way.

the question: What is the derivative of the inverse of any function f(x)

the answer:

1/[f'(y)][/size]

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I will post my work. I will quote exactly what I wrote in a wordpad document to generalize this:

f(x)=x^3+x g(x)=f-1(x)

g(2)=1 g'(2)= ?

-------------------------------------------------

y=x^3+x => x=y^3+y

y^3+y=x is the g(x) in a bad form but correct.

3y^2 dy/dx + dy/dx = 1

dy/dx = 1/(3y^2+1)

-------------------------------------------------

f(x)=ax^1+bx^2+cx^3+dx^4+...+zx^n g(x)=f-1(x)

g(a)=b

g'(a)=?

inverse: x = a f(x)^1 + b f(x)^2 + c f(x)^3 + d f(x)^4 +...+z f(x)^n (1)

g(x)= a f(x)^1 + b f(x)^2 + c f(x)^3 + d f(x)^4 +...+z f(x)^n (2)

g = ay^1 + by^2 + cy^3 + dy^4 +...+zy^n (3)

Range of f(x) is domain of g(x) and domain of g(x) is range of f(x) becasue an inverse function is defined as f(y)=x

1=ay^0 dy/dx + 2by^1 dy/dx + 3cy^2 dy/dx + 4dy^3 dy/dx + nzy^(n-1) dy/dx (1)'

dy/dx = 1/[ay^0+2by^1 + 3cy^2 + 4dy^3 + nzy^(n-1)]

---------------------------------------------------

For a function f defined by f(x)=ax^m+bx^n+cx^p+dx^p...
The dy/dx of the f' is equal to 1/[amy^(m-1)+nby^(n-1)+pcy^(p-1)+pdy^(p-1)]

Where y = the range of f(x) at a point x.

---------------------------------------------------

dy/dx of f-1(x) = 1/[f'(y)]

applying it for a simple equation for verrification: f(x)=x^3+x the one given above.

dy/dx of f-1(x)=1/[3y^2+1] from above and from my formula.

 

[Hurkyl]

Yep, that's the right formula!

Judging from your post, though, you've only proven it for polynomials. Can you figure out how to prove it for an arbitrary differentiable, invertible function? (Hint: it's the same basic idea, but the details are simpler than what you've done with polynomials)

 

[Robokapp]

let's try a simple f(x)=e^x

my formula says that the derivative of inverse should be 1/f'(y)

so 1 /[ d/dx of e^(y)]
1/[e^(y)*(y')] is what I seem to get.

1/(y'f(y)) maybe!

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I can't prove it...I get stuck in a big pile of information andit's too late for that...I'll take a look tomorrow.

What I'll try to see is is 1/[y'f(y)] same as 1/f'(y)

tops cancel. y'f(y)=f'(y)

by Chain Rule the derivative of f(y) is f'(y)*y'

Oh shoot. I chose a bad example, didn't I? e^x and its derivative are the same so I can't tell if my formula is 1/[y'f(y)] or 1/[y'f'(y)]

Well, it's friday night and I'm already dizzy so I'll trust my previews work and the Chain Rule and state that the formula 1/[y'f'(y)] works.

However...it still won't work meaning I'm terribly wrong. I give up...I'll look at it tomorrow!

 

[masudr]

Think about the chain rule.