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Advanced Differentiation

Problem Sets (Non Javascript - Problems | Answers)

Implicit Differentiation

This will only be easy if you can follow the steps. Understanding it is nice too, but why waste your time with something so trivial as actually knowing what it is you are learning. 

In this section, you are taking a derivative, except the issue raised here is that you cannot always isolate y onto one side of your equation, making a different method necessary.

Here is a typical example:

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 y’.

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. 

Logarithmic Differentiation

OK, here’s the deal. I don’t like doing this stuff, so let’s get over with it.

y = x^2x  I explained how to do x^a and a^x 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 a^b 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 = x^2x

ln y = ln x^2x = 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’ = x^2x(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)]

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