# Working With Derivatives

So far in the last section, we figured out how to differentiate, or take the derivative, of a function using limits. However, it is cumbersome to always evaluate to find the derivative. An easier way to find the derivative of a function, say is to multiply the coefficient of x by it's exponent and then subtract one from the exponent. This can be summarized by the equation . So for , we multiply the coefficient of x, understood to be 1, by the exponent which is two, and then subtract 1 from the exponent giving us =2x which is the same answer we got when we took the limit as h goes to zero on the last page.

Let's try another case. Suppose . In this case we apply the above rule to each of the terms which would give us: . Note that the constant 4 in is lost when we differentiate because the exponent on 4 is understood to be 1, and when we subtract one from the exponent when we are differentiating the exponent on 4 goes to 0. Anything to the 0 power is 1.

However, this only works in certain cases when is in the form where 'a' and 'n' are real numbers.

Suppose we had something like: and we wanted to take it's derivative. In this case, we use what is called the product rule which is used when we are taking the derivative and we are multiplying. The product rule looks like: Here f(x) = (x - 2) and g(x) = (2x + 1). f'(x) = 1 and
g'(x) = 2. Substituting these into the rule we have: which simplifies to [f(x)g(x)]' = 4x - 3 which is the derivative of .

That's what we do when we are multiplying. There is another rule called the quotient rule that we use when dividing. It looks like: For an example in the use of the quotient rule let's try: . We have to use the quotient rule because the function is in the form of a fraction. Well we don't have to use the quotient rule; we can use product rule on , but's let's try to use the quotient rule. We are going to need f' and g'. So f'(x) = 2 and g'(x) = 4. Putting these values and f(x) and g(x) into the quotient rule we have: which simplifies to: .

We have one more topic to discuss for this section and then we are finished. So far we have only discussed first derivatives. Second and third derivatives also exists. Some functions have an infinite number of derivatives like the cosine and sine functions, but that's in the trigonometric derivatives section.

Taking second and third derivatives is just like taking the first derivative of function with the exception that the function has been differentiated once already. Let's try to take some second and third derivatives and also use the new notation that is involved. Let's use the function that we worked with above: . We already know . We can now take the second and third derivatives which is just like taking the first derivative: . The dydx notation is read 'the second derivative of y with respect to x squared.' The third derivative is found in a similar manner: .