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Wednesday, November 12, 2008

More of the Chain rule, Natural Logarithms, e! Sugoi!

Previously on AP Calculus 2008...
(Monday, November 10th)

We have 4 rules for derivatives so far:

The Power Rule: d/dx [x^(n)] nx^(n-1)
The Product Rule: d/dx [f(x) * g(x)] = f(x) * g'(x) + g(x) * f'(x)
The Quotient Rule: d/dx [f(x) / g(x)] = [g(x) * f'(x) - g'(x) * f(x)] / [g(x)]^2
The Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g(x)

And now the continuation...

So, we basically continued working with these rules. The topic of the day was The Leibniz notation for the Chain Rule.

(All this is in the book, pages 220 to 222)

We have two differentiable functions, defined thus:

y = f(u) and u = g(x)
Therefore y = f(g(x))

Then we know: dy/du = f'(u) and du/dx = g'(x). If we apply the chain rule...
dy/dx = f'(u) * g'(x) = dy/du * du/dx

Simply put: dy/dx = dy/du * du/dx

So if we put this into an example:

y(u) = (sin3x)^6
u = sin3x

We can apply the chain rule and power rule here to find dy/dx y(u):

dy/du = (6sin3x)^5
dy/du = 6u^5

Now here we start with the "layers" Dr. Eviatar talked about. Theoretically, your function could be so long that there would be functions within functions (within functions, within functions...!), thus having many "layers" of functions. The usefulness of the chain rule is that we can use it to seperate the layers and find the derivative.

So in this example...

dy/du * du/dx = dy/dx 

The thing is, finding the derivative of sin3x (du/dx) isnt simple. So we'll just seperate it into "layers" with the chain rule, again. Let's have another variable, v.

du/dx = dv/dx * du/dv

And have v = 3x
So now, we have an equation within an equation.

u = sinv

du/dv = cosv
dv/dx = 3
du/dx = cosv * 3 = 3cosv = 3cos3x

And finally find the derivative:

dy/dx = 6sin3x^5 * 3cos3x

Arent you glad there were just two layers?

After that, we looked over the question on page 221 about the oil slick. This question is the same as any other derivative question we've looked at, but with different variables.

dr/dt = 3ft/hr
dA/dr = 10ft^2/ft

A = f(r) where r = g(t)  therefore A = f(g(t)). Thus we can apply the chain rule:

dA/dt  = dA/dr * dr/dt = 10ft^2/ft * 3ft/hr = 30ft^2/hr

Simple, right?

And finally, we started into the mysterious and awe inspiring power of e and the natural logarithm.

e^(lnx) = x

We find the derivative of both sides:
d/dx [e^lnx] = d/dx [x]

The derivative of x is 1 because x is a constant. We apply the chain rule to find the derivative of e^lnx:

e^lnx * d/dx(lnx) = 1 

Remember, we stated earlier on that e^lnx = x...

x * d/dx(lnx) = 1

So finally, the derivative of lnx must be:

d/dx [lnx] = 1/x

So now that we know this...

And that concludes yesterdays episode of AP Calculus 2008. Tune in tomorrow for more exciting mathematical developments!

Oh and we had some questions to solve...
Find the derivative of the following and which rule(s) would you use to solve it:

Function | Rule
ln2x | Chain Rule
xlnx | Product Rule
(lnx)^3 | Chain + Power Rule

And some homework I'm sure you all did. So because you all did it, I dont have to write down the question numbers you were supposed to do. Because you already did them.

The next scribe is "Hi Im Justu"

Whoever the heck that is.

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