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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 3

Lesson 8: Calculating higher-order derivatives

# Second derivatives (implicit equations): evaluate derivative

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.F (LO)
,
FUN‑3.F.1 (EK)
,
FUN‑3.F.2 (EK)
Given the first derivative of an implicit equation in x and y, evaluate the second derivative at a certain point. Problem taken from the 2015 AP Calculus AB exam.

## Want to join the conversation?

• How is it apparent that you should use the quotient rule here, and not rewrite the expression to use the product rule?
• The denominator here is complex, so the denominator to the (-1) would be much more complicated, so we just use the quotient rule and then substitute the value of x,y respectively.
• Even if the method is straightforward, the calculation involved is very challenging.
• Does it matter if you calculate the value of the first derivative (1/4 in this case) and then sub it into the second derivative equation? Or is it equally acceptable to sub the equation for the first derivative directly into the second derivative equation and then use the values of x and y to calculate?
• Did you try them both for the problem shown in this video?

What did you find?

Both of those approaches are mathematically equivalent and should give identical results ...
• Why do we put the squared on the d on the top and the squared on the x on the bottom of the derivative operator?
(1 vote)
• dy is called the differential of y; you can think of it as an infinitesimally small change in y. In other approaches to calculus, we define the derivative of y with respect to x as the differential of y divided by the differential of x, dy/dx.

To take the second derivative of y with respect to x, we take the differential of the differential of y, d(dy), and divide it by dx twice. So the second derivative is d(dy)/(dx)².

Then we abbreviate ddy as d²y, and drop the parentheses in the denominator to get d²y/dx².
• This isn't related to the question but why is the second derivative written as d^2y/dx^2?
(1 vote)
• Sal explains it in the first video on Second Derivatives. Basically, d^(2)y/dx^(2) is d/dx(dy/dx). So, the numerator kinda becomes d^(2)y and the denominator becomes dx^(2). Note that you don't multiply the terms, as d^(2)/dx^(2) is an operator just like d/dx. All in all, it's pretty much a convention someone long ago came up with, and we continue using it
• In practice problems for second derivatives, implicit equations, problem 2 (x^3+y^2=24), step 3/5: what manipulation did you do to go from [-6xy+3x^2(-3x^2/2y)]/2y^2 to [-12xy^2+9x^4]/4y^3?
(1 vote)
• So this is equivalent fractions. Remember yr 7 math.

You have just multiplied the numerator and denominator by 2y.
(1 vote)
• When differentiating x^y, the result is ln(x)x^y. Can anybody explain why we do that, or point be to the module where that comes up?
(1 vote)
• I did not get this answer
(1 vote)
• When should we use implicit differentiation or explicit differentiation? Can someone help me out of it...
(1 vote)
• Implicit differentiation is most often used when the equation cannot be made explicit or the differentiation of the explicit equation would be more complex.

For example, the equation y^5 + y^3 = x cannot be manipulated in order to get y by itself on the left side (i.e. there is no explicit form of the equation). So, we have to implicitly differentiate.

An example that could be explicitly differentiated, but probably should not be, is sin(x*y) = cos(x). Solving for y and then differentiation is possible, but it would be tricky. It would be easier to simply implicitly differentiate the equation as it is.
(1 vote)
• Why pug in x=-1 and y=1 into the dy/dx equation and not the "original" equation given (y^3-xy=2)? I'm confused because the question says, "...the coordinates on the curve" and the equation of the curve we have already. Thanks.

-Aviel
(1 vote)
• If you plug (-1, 1) into the equation of the curve, it simplifies to 1+1=2. We already knew it would do that, because this is a point on the curve, so of course plugging a point on the curve into the curve gives a true statement.

The derivative equation gives a relation between these same points on the curve and the slope of the curve, so plugging a point into that equation and solving for dy/dx will tell us the derivative at that point.
(1 vote)
• Why does it not matter for these functions if the given point to evaluate the derivate at does not exist in the function? (One of their practice problems asks me to evaluate the second derivative of y^4-2x = 14 for the point (-2, 1)).