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Course: AP®︎/College Calculus BC>Unit 3

Lesson 8: Calculating higher-order derivatives

Second derivatives (implicit equations): evaluate derivative

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.

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• 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.
• 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.
• 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 ...
• Hello! I have tried https://www.khanacademy.org/math/differential-calculus/dc-chain/dc-second-derivatives/e/second-derivatives-implicit-equations
100 times and i fail every time! How can I improve? Every time I miss something, a minus, an exponent, something. I see it clearly when I see the entire calculation but I'll be damned if I see it when I'm trying. I really must be having some basic problems.
• Do you write out your work on paper? I find its easier to avoid mistakes/notice mistakes when I make them when I write out my work on paper. Also for the ones where you evaluate it at specific number, I find that checking in a calculator helps to make sure you don't make silly arithmetic errors when plugging specific points in. This won't help you get the correct expression in the first place, but it does help avoid some of those silly mistakes. In general, practice helps always, and it's better to not rush through the problems (I struggle with this sometimes)
• Could someone explain in depth what dy/dx is? It seems to show up at random times whenever there is a y, and I don't really understand why
(1 vote)
• dy/dx means the derivative of y (or y'). It's used whenever you take the derivative of an equation. When doing so you substitute y with dy/dx.
• 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
• When doing the practice examples I have a question. The second derivatives that take a derivative of -(x/y) have a quotient rule that doesn't match a product rule answer, why is that? I'm assuming it has something to do with the problem of having two unknowns. Shouldn't the functions f(x)=x and g(y)=y, if the broader function is f(x)/g(x), shouldn't this still be equivalent to f(x)*g(x)^-1?
(1 vote)
• It is important to be clear. If you really want you could use a derivative calculator.

If you could write your solution then I can check it. I would avoid using quotient rule as it is more complicated then product rule.

To answer though product rule/quotient rule are equivalent