3y2 = 2x + x2y
Actually, this one’s a bit tough, I think. Anyway, you should already see that it is impossible to take a regular derivative. y is not a simple thing off to one side, and to isolate y off to one side would be difficult. So we are going to take a straight derivative of each or the 3 terms here, and then try to isolate y’, which is much easier. That way we will have a derivative for the equation.
One important thing to remember when doing implicit differentiation: The derivative of y is y’. The derivative of x is taken in the normal way of a variable, so that the derivative of x is 1, as an example. The reason for this is that y is a function of x, and does not exist without being dependant on the equation on the other side, on the other hand, x exists all along the domain. I mentioned this briefly in my overview of derivatives.
Now that we know this, it should make sense that the derivative of y2 is 2yy’. The reason is that you actually use the chain rule to solve it. x2 becomes 2x *1 because the derivative of the inside function, x, is one. Here the derivative of the inside function, y is y’. So the outside function becomes 2y, and the inside becomes y’, you multiply them together and you have 2yy’. So the derivative of 3y2 is 6yy’. Onto my problem.
3y2 = 2x + x2y 6yy’ = 2 + 2xy + x2y’
I used the product rule to solve the third term.
The general procedure is as follows:
Step 1 – Differentiate both sides, like I just did.
Step 2 – Send all terms with y’ inside to one side
of the equation, and all terms without to the other side.
Step 3 – On the side with y’ terms, factor out the
Step 4 – Divide both sides by the parentheses factor
on the y’ side.
Step 1 gave me 6yy’ = 2 + 2xy + x2y’
Step 2 gives me 6yy’ – x2y’ = 2 + 2xy
Step 3 gives me y’(6y – x2) = 2 + 2xy
Step 4 gives me y’ = (2 + 2xy)/(6y – x2)
See, wasn’t that simple? Here’s one more:
2x2 – y3 = 4xy
4x – 3y2y’ = 4y + 4xy’
4x – 4y = 3y2y’ + 4xy’
4x – 4y = y’(3y2 + 4x)
(4x – 4y)/(3y2 + 4x) = y’
Hope that helps you some. Notice that I used my four steps to the letter. It really works every time.
This one may be a bit more difficult. The ones in the problem sets are even more so.
2y5x = x2y + 14x - 2
10y4y'x + 2y5 = 2xy + x2y' + 14
10y4y'x - x2y' = 2xy - 2y5 + 14
y'(10y4x - x2) = 2xy - 2y5 + 14
y' = (2xy - 2y5 + 14)/(10y4x - x2)
One person emailed me that he first divided everything by x, to make the calculations simpler. His answer came out looking completely different from mine, but I checked, and it turned out that they were really the same. Just letting you know, in case you also have the urge to do things differently.
In this last problem, suppose you wanted to know the derivative of y when x = 1. It's complicated, but look at the last line. you see that y' is a function of both x and y. So if you know the x and the y, you can plug them in and solve. Yet, I've only told you the x. The y you can get because the original equation only has x's and y's, so if you know the x, you can hopefully solve for y. Don't bother though; this problem is not designed to have a clean answer for that.
I will now take this concept, and make it a bit more difficult, but it really shouldn't be a problem if you understand what happened until this point.
What is the second derivative of y2 + y3 = x2 - 52, when y = 2?
2yy' + 3y2y' = 2x
y' = 2x/(2y + 3y2)
Up until here is standard. Now I need the second derivative. I can do it from either of the last two equations.
I will use the first:
y" = [2 - 2(y')2 - 6y(y')2]/[2y + 3y2]
Now you have to plug in a few things to get y". You need y, which is 2. You also need y'. We have to use the equation we made earlier to get this. But that equation has both x's and y's. We only know the y. However, we can use the original equation to get x. If you use the original equation, x will solve to 8, which means y' = 16/16 = 1, and we have all the necessary variables to solve for y". Do it yourself. I'm lazy.
OK, here’s the deal. I don’t like doing this stuff, so let’s get over with it.
y = x2x I explained how to do xa and ax But when you have a variable in both the number and its exponent, the equation is much more difficult. The only way to solve it is to utilize implicit differentiation, in a way.
There is one rule that you need to know. ln (ab) is the same as b ln a. The exponent can be moved to the beginning. This is not taking a derivative; it is simply another way of viewing the same problem. Now we can do that to our problem.
y = x2x
ln y = ln x2x = 2x ln x First I used a natural log on both sides, which is general procedure to solve it, because it will bring down the variable into a situation where we can simply use product rules and such to solve. Implicit differentiation is also necessary.
ln y = 2x ln x
(1/y)y’ = 2ln x + 2x(1/x)
y’/y = 2ln x + 2
y’ = y(2lnx + 2)
y’ = x2x(2lnx + 2)
This generally ends up working slightly different from regular implicit differentiation, but it is very intuitive. Notice how I took the derivative of ln y. The outside function becomes 1/y and the inside becomes y’. One more thing is the last step that I did. I went and replaced the y back with with it is equal to. We don’t do that by implicit differentiation, but here it is simple to, because we just plug in from the original equation.
y = (3x + 2)ln x
This problem is really scary. I don’t know why I wrote it. Here goes.
Ln y = ln [(3x + 2)ln x]
Ln y = ln x ln (3x + 2)
y’/y = ln (3x + 2)/x + 3ln x/(3x + 2)
y’ = y[ln (3x + 2)/x + 3ln x/(3x + 2)]y’ = (3x + 2)ln x[ln (3x + 2)/x + 3ln x/(3x + 2)] Notice that I replaced y back with what it is equal to. We do not do that by a regular implicit differentiation problem.