Derivatives play an important role in real world because they are velocity and acceleration.

A function is displacement (at some t=a, the object is located at x=b).

And when you take the derivative of a function, it becomes velocity (how fast the object moves at some t=a).

Taking the derivative of a derivative (2^nd derivative) will give you acceleration (the rate that the object moves).

*********************DIFFERENTIATION RULES********************

The followings are the several rules you need to know in order to take the derivative of a function:

*NOTE: If f(x) is a function, then derivative can be written f'(x) or

d[f(x)]/dx (Leibniz).

Then derivative of f(x): f'(x) = nx^n-1.

Let h(x) = f(x)g(x) and f'(x) & g'(x) exist.

Then h'(x) = f(x)g'(x) + g(x)f'(x).

Let h(x) = f(x)/g(x) and f'(x) & g'(x) exist.

Then h'(x) = [g(x)f'(x) - f(x)g'(x)]/[g(x)]².

Assume g'(x) & f'(g(x)) exist.

Let h = f ° g implies h = f(g(x)).

Then h'(x) = f'(g(x))g'(x).

You definitely need to know how to use all the rules above if you want to survive in CALCULUS. I am 100% sure and absolutely correct that these 4 rules will be on your exam. So prepare.

Taking derivatives is very easy. Among the 4 rules, most students have difficulties when they use Chain Rule. In fact, it is most complicated rule amongst 4 rules. Most of the text books that I have encountered have one whole section for Chain Rule and the other rules put into one, which indicates that is harder and needs little more thinking than the other rules.

Following examples demonstrate Power Rule, Product Rule, Quotient Rule, and Chain Rule, respectively.

Example 1.2.1:

Take the derivatives of following functions.

1. f(x) = x² + x + 7

Solution: By Power Rule, f'(x) = 2x + 1

2. g(x) = (x²)(5x)

Solution: By Product Rule, g'(x) = (x²)(5) + (2x)(5x)

= 5x² + 10 x² = 15x²

*NOTE: You can easily check by multiply x² and 5x and get 5x³.

Now by Power Rule, g'(x) = 15x².

3. h(x) = 1/x

Solution: By Quotient Rule, h'(x) = [x(0) - 1(1)]/[x]²

= 0 - 1/x²

= -1/x².

4. F(x) = (x² + 1) ²

Above examples look easy and doable. That's because I have chosen very simple examples. Depending on what the functions are, it gets tough.

*******************IMPLICIT DIFFERENTIATION*******************

The examples that you examined above are done in explicit differentiation. We can explicitly differentiate a function only when we can separate independent variable from dependent variable. If you cannot separate them, then you cannot use explicit differentiation. Then what should we do?

We use implicit differentiation. We differentiate both sides of a relation with respect to x and then solve the resulting equation for y'.

Let's examine a simple example that will clarify what implicit differentiation is.

Example 1.2.2:

Find y' by implicit differentiation.

x² + 2x + xy = 0

Solution: Can you get y in terms of x? Yes, y = -(x² + 2x)/x = -(x + 2)

Example 1.2.2 gives y' in terms of x (in this case, a constant). That's because the given relation of y can determined. When differentiating implicitly, you will get y' in terms of both x and y. And you are going leave it the way it is since you can't find y in the first place.

If you have any problems, send me an e-mail.