Implicit differentiation

  • Thread starter BOAS
  • Start date
  • #1
555
19

Homework Statement



Hello,

I missed the class where we were introduced to implicit differentiation so have been catching up this evening. I think I have it, but please could you check my working? Thanks!

Find the derivative of y2 = 2x + 1

[itex]\frac{d}{dx}[/itex][itex]([f(x)]^{2})[/itex] = [itex]\frac{d}{dx}[/itex]([2x]) + [itex]\frac{d}{dx}[/itex]([1])


[itex]\frac{d}{dx}[/itex][itex]([f(x)]^{2})[/itex] = [2f(x)]f'(x)

[itex]\frac{d}{dx}[/itex]([2x]) = 2

[itex]\frac{d}{dx}[/itex]([1]) = 0

So my answer is;

2y[itex]\frac{dy}{dx}[/itex] = 2

Is this correct?

I suppose i'm a bit confused about how I should leave my answer...
 

Answers and Replies

  • #2
Qube
Gold Member
468
1
Yes your answer is correct but to finalize your solution remember what you're solving for. When you normally differentiate something, such as

2x + 1 = f(x).

You get 2(dx/dx) + 0 = dy/dx, and the dx terms "cancel out" and you get 2 = dy/dx.

Does this help?
 
  • #3
555
19
Yes, I see.

It is strange to leave it in the form I did when the whole point of taking the derivative is to see the function of dy/dx.

Thanks
 
  • #4
35,124
6,866

Homework Statement



Hello,

I missed the class where we were introduced to implicit differentiation so have been catching up this evening. I think I have it, but please could you check my working? Thanks!

Find the derivative of y2 = 2x + 1

[itex]\frac{d}{dx}[/itex][itex]([f(x)]^{2})[/itex] = [itex]\frac{d}{dx}[/itex]([2x]) + [itex]\frac{d}{dx}[/itex]([1])


[itex]\frac{d}{dx}[/itex][itex]([f(x)]^{2})[/itex] = [2f(x)]f'(x)

[itex]\frac{d}{dx}[/itex]([2x]) = 2

[itex]\frac{d}{dx}[/itex]([1]) = 0

So my answer is;

2y[itex]\frac{dy}{dx}[/itex] = 2

Is this correct?

I suppose i'm a bit confused about how I should leave my answer...
You already have your answer, from Qube's post, so I'll add some comments about how you wrote your work.

This is what I would do.
y2 = 2x + 1
##\Rightarrow## d/dx(y2) = d/dx(2x + 1)
##\Rightarrow## 2y * dy/dx = 2
##\Rightarrow## dy/dx = 2/(2y) = 1/y = ##\frac{1}{±\sqrt{2x + 1}}##

I don't see any advantage of introducing f(x), so it doesn't appear in my work. Also, each line above is an equation that flows logically from the equation in the preceding line. When you have a line that represents only the derivative of one side of the equation, you lose the flow in the logic.
 
  • #5
555
19
Hi Mark,

thanks for replying, it's always useful to see faster methods of reaching the solution. On the grading scheme we're given when our work is handed back, there is a section commenting on how much time our method would cost us in an exam.

I agree with you that I can't see the advantage of saying y = f(x), but this is what is done in my maths text book.
 
  • #6
35,124
6,866
I agree with you that I can't see the advantage of saying y = f(x), but this is what is done in my maths text book.
If the function is defined as f(x) = <whatever> then it makes sense, but if you're given an equation that involves only x and y, then it doesn't make sense to me to bring in f(x). In any case, y is not a function of x in this problem - for each x value other than -1/2 there are two y values.

That was kind of a minor point, though. What I think is more important is doing your work so that it flows logically.
 

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