Find the Derivative of y2 = 2x + 1How can I find the derivative of y2 = 2x + 1?

In summary: Thanks for your input.In summary, Mark missed the class on implicit differentiation and was trying to catch up by doing some work tonight. His work consisted of finding the derivative of y2 = 2x + 1, which he found to be \frac{d}{dx}([f(x)]^{2}) = \frac{d}{dx}([2x]) + \frac{d}{dx}([1]). When he tried to finalize his solution, he realized that he didn't understand what he was solving for, and was confused about how to leave his answer. He was helped by his classmate who explained that when you differentiate something, like 2x + 1 = f(x), you get 2
  • #1
BOAS
552
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...
 
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  • #2
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
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
BOAS said:

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
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 textbook.
 
  • #6
BOAS said:
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 textbook.
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.
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly expressed in terms of one variable. This can be used when the relationship between variables is not easily solved for one variable.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation involves finding the derivative of a function that is written explicitly in terms of one variable, while implicit differentiation involves finding the derivative of a function that is not written explicitly in terms of one variable.

3. When is implicit differentiation used?

Implicit differentiation is used when the relationship between variables is not easily solved for one variable, or when a function is complex and difficult to differentiate using traditional methods.

4. What are the steps for implicit differentiation?

The steps for implicit differentiation are:1. Differentiate both sides of the equation with respect to the variable of interest.2. For terms with multiple variables, use the product rule, chain rule, or quotient rule as needed.3. Collect all terms with the derivative of the variable on one side of the equation.4. Solve for the derivative.

5. What are the applications of implicit differentiation?

Implicit differentiation has many applications in physics, engineering, economics, and other fields. It can be used to find rates of change, find maximum and minimum values, and analyze curves and surfaces in three-dimensional space.

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