AP®︎ Calculus BC (2017 edition)
- Implicit differentiation
- Worked example: Implicit differentiation
- Worked example: Evaluating derivative with implicit differentiation
- Implicit differentiation
- Differentiating related functions intro
- Worked example: Differentiating related functions
- Differentiate related functions
- Showing explicit and implicit differentiation give same result
- Implicit differentiation review
Review your implicit differentiation skills and use them to solve problems.
How do I perform implicit differentiation?
In implicit differentiation, we differentiate each side of an equation with two variables (usually and ) by treating one of the variables as a function of the other. This calls for using the chain rule.
Let's differentiate for example. Here, we treat as an implicit function of .
Notice that the derivative of is and not simply . This is because we treat as a function of .
Want a deeper explanation of implicit differentiation? Check out this video.
Check your understanding
Want to try more problems like this? Check out this exercise.
Want to join the conversation?
- Hi !
i dont understand why in PROBLEM 1 :
2x + y + *bold*x dy....
why x dy
thank you!(18 votes)
- It's because of the product rule.
x*y differentiate into (1 (from differentiating the x))* (y) + (x) * (dy/dx (from differentiating the y)) = y + x*dy/dx(35 votes)
- Hi everyone, I have a quick question. We use the chain rule to differentiate "y^2" because we treat variable y as a function of x. However, when we have simple "y", we do not apply the chain rule and just express it as dy/dx. What is the difference between y^2 and y? Why to use chain rule in first case and not in the second one like 1(y(x))*dy/dx?(7 votes)
- We already know how to represent the derivative of y with respect to x: dy/dx, which is the thing we wish to find - in terms of x and y.
y² is a function of x AND of y.
Whenever we have a function of y we need to use the chain rule:
d/dx [ f(y) ] = d/dy [ f(y) ] · dy/dx
If it makes you feel easier we could say a 'simple *y"' is the identity function: f(y) = y.
Then d/dx [ f(y) ] = d/dy [ f(y) ] · dy/dx = dy/dy · dy/dx = 1 · dy/dx(17 votes)
- For problem 1, shouldn't it simplify to (-2x-y)/(x+3y^2)(5 votes)
- Yes it could, because (-2x-y) = -(2x+y). So you just distributed the minus one into the numerator, while it's left outside in the solution. It could also have been distributed into the denominator, since -1 = 1/-1.(18 votes)
- Why we can take derivative with respect to x or y both sides during implicit differentiation ?(2 votes)
- You learn in algebra that you can perform the same operation to both sides of an equation and the equation will still hold true. Taking a derivative just happens to be one such operation.(11 votes)
- Find the equation of the tangent line to the graph of the following equation at the point (-1,2) Implicit Differentiation
x^2 y - y^3 = 6x(0 votes)
- Hi everyone!
Do you happen to know any tricks and tips for solving derivatives and limits? Especially for implicit differentiation? I just don't like how long the process is taking me because I am a bit slow at writing and for our exams we have to write and it is time consuming.
Any help is much appreciated. Thank you!(2 votes)
- A short cut for implicit differentiation is using the partial derivative (
∂/∂x). When you use the partial derivative, you treat all the variables, except the one you are differentiating with respect to, like a constant. For example
∂/∂x [2xy + y^2] = 2y. In this case, y is treated as a constant. Here is another example:
∂/∂y [2xy + y^2] = 2x + 2y. In this case, x is treated as the constant.
dy/dx = - [∂/∂x] / [∂/∂y]This is a shortcut to implicit differentiation.
Partial derivatives are formally covered in multivariable calculus.
Even though this is a multivariate topic, this method applies to single variable implicit differentiation because you are setting the output to be constant.
Hope this helps!(7 votes)
- Am I allowed to simplify an equation before doing implicit differentiation? Here is the question I was stuck on: y^2 = (x-1)/(x+1). When I attempt implicit differentiation the way it is and compare the answer to when I simplify the equation to (x+1)y^2= x-1, I got different answers.(4 votes)
- Excellent question!
Implicitly differentiating the original equation eventually yields dy/dx = 1/(y(x+1)^2).
Implicitly differentiating the simplified equation eventually yields dy/dx = (1-y^2)/(2y(x+1)).
So we compare 1/(y(x+1)^2) to (1-y^2)/(2y(x+1)), using y^2 = (x-1)/(x+1).
So the answers are really the same! The point is that the original equation or an equivalent form of this equation must be considered when comparing answers.
Have a blessed, wonderful day!(2 votes)
- I have a question is that if I get a function e^(x/y)=x-y
can I do In both side which gives me x/y=In(x-y) in order to find the derivative?
or I have to d/dx (e^(x/y))=d/dx(x-y)(2 votes)
- I keep forgetting and recognizing that dy/dx(xy)=x'y+xy'. What lessons or video would be good for learning to recognize patterns in math like these, specifically ones having to do with Calculus?(2 votes)
Here's a summary:
------- (df * g)
f | | | (f * dg)
d/dx(fg) = f' g + f g'(2 votes)
- do you treat dy/dx like y? For example, if y is cubed and muliplied by x, is dy/dx?(2 votes)
- Hopefully this helps. Let's say y=x^2 so then cubing it and multiplying ti by x would look like this. xy^3 = x(x^2)^3 = x^7
Now, let's take the derivative of y and this changed y.
dy/dx = 2x
d(xy^3)/dx = 7x^6
Now let's see if we do like you said and cube dy/dx and multiply it by x.
x(2x)^3 = 8x^4
Let me know if that's not what you meant.(2 votes)