Main content
AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 3
Lesson 8: Calculating higher-order derivativesSecond 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.
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?(11 votes)
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.(16 votes)
Even if the method is straightforward, the calculation involved is very challenging.(13 votes)
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?(0 votes)
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 ...(6 votes)
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.(3 votes)
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(2 votes)
- 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)
Can someone show this using the product rule? I know it is more tedious but I want to see how it achieves the same result.
So far I have y(3y^2-x)^-1
Then dy/dx(y)*(3y^2-x)+ y*dy/dx(3y^2-x)^-1
Then (1/4)1*4 + 1*dy/dx(3y^2-x)^-1
Then 1+dy/dx(3y^2-x)^-1
Then using the chain rule >> 1+ (dy/dx(-(3y^2-x)^-2* (dy/dx(6y-1))
Mainly where I'm stuck is where to input the dy/dx into the chain rule section. Once I got -5/16, then 1/128. I'm not getting the same result...(1 vote)- The fractions get really ugly (and unreadable) so I recommend you render this in LaTeX so it'll look better.
If you have $y(3y^{2} - x)^{-1}$, on differentiating using the product and chain rule, we get $y(-(3y^{2}-x)^{-2})(6y(\frac{dy}{dx})-1) + (3y^{2}-x)^{-1}(\frac{dy}{dx})$. Now, from the already given equation of $\frac{dy}{dx}$, we can find $\frac{dy}{dx} = \frac{1}{4}$. Plus, we have $x = -1$ and $y = 1$. So, on plugging all of this in, we get $(1)(-(3(1)^{2}-(-1))^{-2})(6(1)(\frac{1}{4})-1) + (3(1)^{2}-(-1))^{-1}(\frac{1}{4})$. This simplifies to $(\frac{-1}{16})(\frac{6}{4}-1) + (\frac{1}{4})(\frac{1}{4})$, which is $(\frac{-1}{32}) + (\frac{1}{16})$ which is $\frac{1}{32}$. So, as you can see, the product rule works too!(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)
Video transcript
- [Instructor] So we have a question here from the 2015 AP Calculus AB test, and it says consider the
curve given by the equation y to the third minus x y is equal to two. It can be shown that the
first derivative of y with respect to x is equal to that. So they solve that for us. And then part c of it,
I skipped parts a and b for the sake of this video. Evaluate the second derivative
of y with respect to x at the point on the curve
where x equal negative one and y is equal to one. So pause this video and
see if you can do that. All right now let's do it together. And so let me just first write
down the first derivative. So dy, derivative of y, with respect to x is equal to y over three y squared minus x. Well if we're concerning ourselves with the second derivative, well then we wanna take the
derivative with respect to x of both sides of this. So let's just do that. Do the derivative operator on both sides right over here. Now on the left hand side, we of course are going to get
the second derivative of y, with respect to x. But what do we get on the right hand side? And there's multiple
ways to approach this. But for something like this, the quotient rule probably
is the best way to tackle it. I sometimes complain
about the quotient rule saying hey, it's just a
variation of the product rule. But it's actually quite
useful in something like this. We just have to remind ourselves that this is going to be equal to the derivative of the numerator with respect to x. And so that's just going
to be derivative of y with respect to x times the denominator, three y squared minus x Minus the numerator, y, times the derivative of the
denominator with respect to x. Well, what's the derivative of this denominator with respect to x? Well the derivative of three
y squared with respect to x that's going to be the
derivative of three y squared with respect to y, which is just going to be six y. I'm just using the power rule there. Times the derivative
of y with respect to x. All I did just now is I
took the derivative of that with respect to x. Which is derivative of
that with respect to y times the derivative
of y with respect to x. Comes straight out of the chain rule. Minus the derivative of
this with respect to x, which is just going to be equal to one. All of that over, remember, we're in the middle of the
quotient rule right over here. All of that over the denominator squared. All of that over three y
squared minus x, squared. Now lucky for us, they want us to evaluate this at a point, as opposed to have to do a bunch of algebraic simplification here. So we can say, when, let me do it over here. So when, I'll it right here. When x is equal to negative one, and y is equal to one, well first of all, what's the, what's dy dx going to be? The derivative of y with respect to x, let me scroll down a little bit, so we have a little bit more space. The derivative of y with respect to x is going to be equal to one
over three times one squared, which is just three, minus negative one. So that's just going be plus one. It's going to be equal to 1/4. And so this whole expression over here, so I can write the second
derivative of y with respect to x is going to be equal to, well we know that that's
going to be equal to 1/4. 1/4 times three times one squared, which is just three, minus negative one, so plus one, minus one, so I'll just leave that minus out there, times six times one, times 1/4. Let me just write it out. Six, six times one times 1/4 minus one all of that over let's see this is going to be three times y squared, y is one, so is just gonna be three, three minus negative one, so plus one, squared. Now what is this going to be? And this is just
simplifying something here. 1/4 times four that's going to simplify to one. And let's see, this is
going to be 1 1/2 minus one so that's going to be 1/2. And then we're going to be
have all of that over 16. And so this is going to be equal to, we get a mini drum roll here, this is going to be
equal to one minus 1/2, which is equal to 1/2. Over 16, which is the same thing as one over 32. And we are done.