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Complex Derivatives

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Chat about Calculus

To make my point in this section, I have chosen to use the only convenient written method of simplifying the parts of a function, which is to break the function up into u's and v's. If you are not accustomed to this notation, it will be a bit of a turn off, and I'm sorry. However, it really is quite simple to understand, if you are willing to put in an extra couple of seconds. 

An example: f(x) = 10x2e2x. If I want, I can call this a product of more than one function. There are a thousand ways to break it up, but one is readily apparent. 
This function is 10x2 multiplied by e2x, so I will break it up into these two pieces.


u(x) = 10x2 and v(x) = e2x 
f(x) = 10x2e2x = 10x2 * e2x = u(x) * v(x)

The Product Rule

If a function is of the form f(x) = u(x)*v(x) then you cannot take a straight derivative. An example is y = (x2)*(ex).  We only know how to take the derivative of u(x), and the derivative of v(x), but the combined function is a new entity, and uses a formula based on the derivatives of u and v. The way to do this is with the Product Rule. This is:

f = uv
f = (uv) = uv + uv

Consider your equation as two equations multiplied by each other. Call one u and the other v, and uv = y, making (uv)' = y'. You then plug into the four parts of the equation above their respective items, and solve algebraically for y'. Watch:

y = x2ex    

I first determine what u and v are, and then what u' and v' are. The next step is to simply plug these four units into the equation above, which you will have to memorize. It's not that hard, really.

u = x2    v = ex    u' = 2x    v' = ex
y' = (2x)(ex) + (x2)(ex) = xex(2 + x)

y = 3x4ln 7    y = 12x3ln 7 + 3x4 * 0 = 12 * x3 * ln 7    The derivative of ln 7 is 0, because ln 7 is just a number. (1.946) So the second half is 0, and can be ignored. The other way of doing it is by grouping ln 7 with the 3 in front, because they are all numbers, and they can act as a coefficient. I will transform it in the next equation.

y = (3 ln 7)x4      y' = (3 ln 7)4x3 = 12 * ln 7 * x3 

y = 2xx2     u = 2x,   v = x2,   y' = (2)(x2) + (2x)(2x) = 2x2 + 4x2 = 6x2 

A simpler way of doing that last one would be to shorten the original equation to 2x3 and then the derivative is a simple 6x2.

y = 3xx5     u = 3x,  v = x5,   y' = (3xln 3)(x5) + (3x)(5x4) = 3xx4(xln 3 + 5) 

y = x2ln x    u = x2,  v = ln x,   y' = (2xln x) + (x2 * 1/x) = x(2 ln x + 1)


The Quotient Rule

The quotient rule is very similar to the product rule in its purpose and implementation. It deals with a function that is too complex to take a simple derivative, only this time the function is of the form f(x) = u(x)/v(x). An example is y = (x2)/(ex). The only way to take it is by the Quotient Rule.  This is:

(u/v) = (uv uv)/(v2)

Be careful not to mix up the u and the v here. They are two different things the u is the numerator and the v is the denominator, of the original equation.

y = x2/ex   y = (2xex x2ex)/e2x  Firstly, you remember that the derivative of ex is ex. Secondly, as explained in the Intro page, when a function is set to a power, that exponent should be multiplied with its current exp, so (ex)2 = ex*2 = e2x Another way of figuring that out is by realizing what (ex)2 is. Its ex * ex and when multiplying the same number to exponents, just add the exponents. x + x = 2x.  Third this equation can be factored out. Its a bit difficult to see written the way it is, but you can pull an ex out of every term, and cancel the top and bottom, getting x(2 x)/ex.

y = (ln x)/x    u = ln x, v = x,   y' = [(1/x)(x) - (ln x)(1)]/x2 = (1 - ln x)/x2 


The Chain Rule

This is a bit more complicated than the previous two rules. You have to realize what is on the inside and what is on the outside. The method for solving every chain rule problem will be by locating the outside and inside function. The only way to teach this is by representing functions with u's and v's, so you must make an effort to know what I mean by each letter. 

I will structure all problems as such:

v = outside function

u = inside function

y' = v'u' (the derivative equals the product of the derivatives of the two functions)

There are the four general types of outer functions found in chain rules. These are the outside (overall) functions. The u in each represents an inside function, whatever it is.

v = u2    v = eu   v = ln u    v = sin u

y = (2x)2  One way to solve this would be to make it 4x2, using algebra, and the derivative is obviously 8x. But Im going to make it complicated. I want to take the derivative using the Chain Rule. The Chain Rule says that in order to take the derivative you must multiply the derivative of the outside function by the derivative of the inside function.

Step 1 Find the inside function. Notice the four function types I showed. One is v = u2 In this case, the inside function is u = 2x. Remember to set the inside to u.

Step 2 Find the outside function. If u is 2x, then the right side goes from (2x)2 to u2 . The outside function is now v = u2. Remember to set the outside to v.

Step 3 Multiply the derivative of the outside by the derivative of the inside. v = 2u and u = 2 so 2*2u = 4u.

Step 4 Replace u with its original value (the inside function). Here that was 2x, so the answer is 4(2x) = 8x

The idea of a chain rule is always to strip down the function into a useable one. Here are other places you may find replacing the inside function useful. Try to put any of these through the four steps I just outlined.

y = e3x + 2 = eu  (and u = 3x + 2)         y = eu(3) = 3e3x + 2

y = ln (x3 ex) = ln u  (and u = x3 ex)       y = (1/u)(3x2 ex) = (3x2 ex)/(x3 ex)

y = sin (3x) = sin u (and u = 3x)    y = (cos u)(3) = 3cos 3x

I havent explained much about trig functions yet, but I will get to it in the next topic.

ln [(7x + 2)3   This one requires a chain rule inside a chain rule. 

[1/(7x + 2)3] * 3(7x + 2)2 * 7 = 21/(7x + 2)


Nasty Cross Breeds

These are usually a pain. Some equations use two, or even three, rules together. With experience, though, they become natural. Ill give a few, but Im really not in the mood of making complex equations, and I think I put some in the problem sets anyway. Its too stuffy in here.

y = exlnx = eu  (and u = xlnx) This seems like the others at first, but realize that you have to take the derivative of the outside, and the derivative of the inside. The derivative of the outsides easy: eu. However, to find the derivative of the inside you must use the product rule. It becomes 1(ln x) + x (1/x) = ln x + x/x = ln x + 1. So the final answer is (ln x + 1)(exlnx) because I replaced the u back with a (xln x).

y = (ex3x + 2)4 = u4 (and u = ex3x + 2)   y' = 4u3(3ex + ex3x) = 4(ex3x + 2)3(3ex + ex3x)

That one was a bit complicated. The inside function was clear, but to find the derivative, once again, you needed to use the product rule.

y = sin3 4x = (sin 4x)3 = u3     u = sin w, w = 4x,   y' = 3u2 * cos w * 4 = 12(sin2 4x)(cos 4x)

This equation used a chain rule inside a chain rule. The outside function was u3, the inside was sin 4x. We know how to do the derivative of the outside function, but to do the inside we need yet another chain rule, which I marked w. Then I multiplied all 3, but an easier way may be to first figure out the derivative of sin 4x individually using its own chain, and then plugging that in to the regular function when you have to put in the derivative of the outside times the derivative of the inside.

Another way to solve a chain inside a chain, one which I prefer by far, is just to multiply the derivative of all three links. Or four, if the innermost function has a chain inside it. For the problem ln (x2 + 4x)5 you see three functions: ln u, w5, and x2 + 4x. Multiply the derivatives of these three functions, and replace back the u and w, and the answer is y' = (2x + 4)5(x2 + 4x)4/(x2 + 4x)5 = (10x + 20)/(x2 + 4x).


Tangent Lines

A tangent Line means the line that will brush against the curve at the point you designate. The formula for this is extraordinarily simple, yet some people just can't figure it out. By the way, I'm really sorry about the drawing, I'm not exactly a pro at that.

(y y1) = slope * (x x1)


I will now explain each piece of this formula and how to get it.

y remains as y. This y is the y that remains as the y in the equation in the tangent line. Similarly, the x remains as an x. The y1 and the x1 are replaced with their respective coordinates from the point you are trying to get the tangent line for. For the coordinate (3, 4) the x1 is replaced with 3 and the y1 is replaced with 4. The last part of the formula, the slope, is found using derivatives. What you must do is take the derivative of the equation, and plug in the point of x from your coordinate. In (3, 4) you will take the derivative and plug 3 into x, and solve. The answer will be the slope. Plug it in, and isolate y. You're done. Let me illustrate.

What is the tangent line to the equation y = x2 + 2 at the point x = 2?

Step 1 - Find the full coordinate and plug it in. to the formula.

y = x2 + 2     y = 4 + 2     y = 6

So the coordinate is (2, 6). I will now plug it in to (y y1) = slope * (x x1).

(y - 6) = slope * (x - 2)

Step 2 - Find the derivative of the original function.

y = x2 + 2      y' = 2x

Step 3 - Find the slope at your coordinate by plugging in x, and plug that slope into the tangent line equation.

y' = 2x = 4

(y - 6) = slope (x - 2)   becomes   (y - 6) = 4 (x - 2)

Step 4 - Isolate y in the tangent equation.

(y - 6) = 4 (x - 2)       y - 6 = 4x - 8      y = 4x - 2

And that is your final answer. Easy as pie.

What is the tangent line of the equation y = 2x3 - x + 5 when x = 1?

Coordinates:  y = 2 - 1 + 5 = 6

Coordinates are (1, 6)

Derivative: y' = 6x2 - 1

When x = 1,  6x2 - 1 = 6 - 1 = 5

Equation for tangent line:

(y - 6) = 5(x - 1)

y = 5x + 1

Find a tangent line(s) to the curve y= x2 that passes through the point (1,-4).

This is a problem which is rather difficult, and I believe that most teachers won't ask it. If you don't understand what I'm doing, ignore it.

Given: y=x2 and y'=2x

For any point (a,b) on the graph, the formula for tangent is then (y-b) = 2a(x-a).

Because our curve is y=x2, b is a2, so that becomes 
y = 2ax - 2a2 + a2 
-->   y = 2ax - a2

We now have a tangent line for point (a, a2). This works for every 'a' on the
curve. So, when does this line go through (1, -4)? Why don't we plug it in:

-4 = 2*a*1 - a2

a2 - 2a - 4 = 0 

You can see that this has two possible solutions of a. That means that there are two tangent lines to the curve (from two points, obviously) that pass through the point (1,-4). Using quadratic formula, you get 1+/-√5. It works. I tried 1+√5 by plugging this value for 'a' into the equation for our tangent line. It does create a line which is clearly both tangent to the graph at the point (1+√5, 6+2√5), and goes through (1,-4), which you can see by plugging it all in. In case you're having trouble with the tedious work, the final tangent line equation I got was:

y = (2 + 2√5)x - 6 - 2√5   Plug x=1 into this equation, and y is -4. 


Something interesting that will help convince you this all makes sense

I don't know if you've tried, but you can turn any function into a function needing product, quotient, or any other rule. Let's try x2.

y = x2 = x * x 
u = x    v = x
y' = u'v + uv' = 1(x) + x(1) = 2x

y = x2 = x3/x
u = x3    v = x
y' = (u'v - uv')/v2 = (3x2(x) - x3(1))/x2 = 2x3/x2 = 2x

y = x2 = (x)2
u = x    v = u2
y' = v'u' = 2u(1) = 2u = 2x

I hope this helps you understand how flexible these calculus methods are. It's also kind of fun if you ask me.

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Copyright 2004 Bruce