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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 3
Lesson 3: Implicit differentiation- Implicit differentiation
- Worked example: Implicit differentiation
- Worked example: Evaluating derivative with implicit differentiation
- Implicit differentiation
- Showing explicit and implicit differentiation give same result
- Implicit differentiation review
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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
Let's differentiate
Notice that the derivative of
Want a deeper explanation of implicit differentiation? Check out this video.
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Hi !
i dont understand why in PROBLEM 1 :
2x + y + *bold*x dy....
dx
why x dy
dx
thank you!(21 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(44 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?(6 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(20 votes)
- For problem 1, shouldn't it simplify to (-2x-y)/(x+3y^2)(4 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.(20 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.(12 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)- => y(2x) + (x^2)(dy/dx) - 3(y^2)(dy/dx) = 6
=> dy/dx = (6 - 2xy) / (x^2 - 3y^2)(1 vote)
- 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).
(1-y^2)/(2y(x+1))
= (1-(x-1)/(x+1))/(2y(x+1))
= (2/(x+1))/(2y(x+1))
= 1/(y(x+1)^2).
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!(5 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)
Hi guys! I have a question: When does the pain end? (Asking for a friend) Thanks!(4 votes)- When you understand it(2 votes)
- Guys.. I have a question that's been bugging me:
Find the derivative function dy/dx of
x³y³-3x²y+3y²=5xy
Trying to confirm my amswer.(3 votes)- What answer did you get?(2 votes)
does someone know the derivative of cosx=2y-xy?(2 votes)
differentiating both sides with respect to x:
-sinx=2(dy/dx)-(y +x(dy/dx))(3 votes)