Derivative Rules

Derivative Rules

The following rules are used to evaluate the derivatives of functions. As with all skills in Mathematics, proficiency results from experience gained from doing problems. The demonstrations/proofs of these rules are done in terms of the limits of their difference quotients and are provided at the end of this document. For these rules we will use x as the independent variable. 1. constants For any constant k: Example: Example: Example: A expression that is constant means that it does not change with respect to the independent variable. This means it has no rate of change with respect to that variable. Its derivative is therefore zero. In the last example above, we could replace k with a variable, say w, and if this variable w does not depend on x as its independent variable then it remains constant with respect to any change x, and its derivative with respect to x would therefore be zero as well. 2. powers For n ≠0, “Multiply by the exponent n then subtract one from the exponent.” Example: Example: Example: note: 3. constant factors For any function f(x) differentiable at x and with the constant factor a ≠ 0, We can always factor out a constant multiplier prior to evaluating the derivative of the function. Example: Example: Example: 4. sum of functions   For functions f(x) and g(x), both differentiable at x “The derivative of the sum is the sum of the derivatives.” Example: Example: 5. product rule For functions f(x) and g(x), both differentiable at x Example: f(x) = x2. g(x) = x. Example: f(x) = (x2 + 3). g(x) = (x/SUP> – 12). I colored each to more easily follow the steps. Here I use the sum of functions rule. I included the derivatives of the constants 12 and 3 for completeness. One could carry out the binomial multiplication then differentiate using the sum of functions rule. This product rule is necessary for products of non-polynomial functions. We'll see this later when we discuss the derivatives of special functions. 6. reciprocal rule If f(x) is differentiable at x and f(x) ≠ 0 then This follows directly from the quotient rule (below) as a special case. Example: f(x) = x. note that 1/x is the same as xB>-1 and using the power rule we get the same result: -x-2 Example: f(x) = x2 + 4. It is OK not to multiply out the denominator in results such as this one. 7. quotient rule For functions f(x) and g(x), both differentiable at x and g(x) ≠ 0 at x then Example: The reciprocal rule is this special case of the quotient rule. f(x) is set to 1. Example: f(x) is (x+1). g(x) is (x2 – 2). I colored each to show this step. The difficulty with this rule and particularly this example is keeping track of the binomials. Here I evaluated each derivative; I left out the derivatives of the constants 1 and 2, they're zero. Here I reordered the terms to write the polynomial in the conventional way: in decreasing exponents. 7. Chain Rule (composite functions) If f(x) and g(x) are differentiable at x then their composition f(g(x)) has the derivative Explanation: The rate of change with respect to x of the function f will depend on the rate of change of the function g with respect to x. In particular, the rate of change of f with respect to the value of g at x will be the derivative of f(g( x)). Now we need to account for the rate of change of g with respect to x. So we multiply the result for f with the derivative of g with respect to x. Example: g(x) = x2 + 4 and f(x) = x3 So, f(g(x)) = (x2 + 4)3 We use the power rule on the exponent 3 first. And we multiply by the derivative of the binomial. Now we simplify.   Example: f(x) = (x)-2 and g(x) = 2x - 1/x So, f(g(x)) = (2x – 1/x)-2 Notice how I left the negative exponent rather then writing this expression as the denominator of a fraction. I proceed as in the previous example. the exponent -2 becomes -2 -1 = -3. the minus sign comes with it as the multiplier. I evaluate the derivatives of the second term as usual. next to last step: note the second minus sign from the derivative of the reciprocal 1/x.   Chain rule, part 2 The dx in our previous examples names the independent variable involved in the derivative. It is called a differential and is used in advanced Calculus courses. The dependent variable can be any variable we choose and by convention, in general, we choose y. Another notation to represent the derivative in terms of the dependent variable is: This notation is convenient for the chain rule so we're saying that: One is equivalent to the other.   Now suppose we have the functions y = f(u) and u = g(x) . We can write y in terms of x by the composite of these functions like so: y = f(g(x)). By the chain rule and this new notation we have   On the left hand side we note that f(g( x)) is a function in x. So we use the notation . On the right hand side we note that g(x) has the dependent variable u, so we see that f is a function in terms of u as well. So we write its derivative in terms of u, not x and use the notation . Finally, the dependent variable for g is u so g's derivative is written using the notation . We pointed out that the derivative is the rate of change of the dependent variable with respect to the independent variable. So we can interpret   to mean the rate of change of y with respect to x is the rate of change of y with respect to u multiplied by the rate of change of u with respect to x. Explanation: if u changes 5 times faster than x, and y changes twice as fast as u then y must be changing 2  5 = 10 times as fast as x. If we have more composite functions then we extend the 'chain' of derivatives: let y = f(u), u = g(v), v = h(x) then   If you think back to unit analysis then the terms on the right hand side contain factors that “cancel” leaving the left hand side of the equation. Keep in mind though, these factors are not units, they're differentials, yet this analogy is useful setting up the chain of derivatives.   Example: y = 3u + 1 u = x2 Direct substitution of u into y then using the power rule to get y = 3x2 + 1 and then finding the derivative is just as valid.   Example: y = 6u3 u = x2   important step here: We substitute x back in for u to get an expression in terms of x. Example: all steps shown here substitute x back in for u remove the negative exponent Derivations (advanced): Recall the limit of the difference quotient that defines the derived function:     #1: constants f(x) = k, then f(x+h) = k and the difference quotient becomes f(x+h) = k since for all x = x+h f(x) = k.     #2 powers   f(x) = x n and f(x+h) = (x+h)n Step 1: We need to expand the binomial and we do this using the binomial theorem (The binomial theorem is at end of this document.) Step 2: Notice that the first and last terms add to zero. Step 3: We divide the denominator 'h' into each term since it is a factor in every term. Step 4: Every term except the first has h to some power as a factor. We know that for all of these h factors as h approaches zero so do these factors. At the limit these factors vanish. All that is left is the first term nxn-1. n is a constant as is xn-1 since this limit is in terms of h, not x. Step 5: We are left with required result. . And we have a1x n-1 h = nx n-1h The remaining terms have the factor h raised to a power greater than one. So, when dividing this numerator by the denominator, h, an h remains as a factor in each of these remaining terms. And, as we evaluate the limit as h approaches zero, every one of these terms approach zero as well because of this factor h. All that remains is nxn-1h and dividing this term by the denominator h we are left with nxn-1which is unaffected by h as h approaches zero. #3 constant factors A constant in a limit can be factored outside the limit.   #4 Sum of Functions Here we treat the sum of functions as a function itself, substitute that function into the limit expression then collect terms to form the two difference quotients.     #5 The Product Rule In this derivation we need to add and subtract the terms in red. This strategy allows us to introduce the necessary terms to collect the appropriate terms we need to factor into the two difference quotients to satisfy the rule. This insight came with some trial and error and is typical of many proofs/demonstrations in Mathematics. The colors green and blue are used to indicate the terms combined together to arrive at the result.       #6 Reciprocal Rule This derivation follows from direct application. #7 Quotient Rule In this derivation we combine fractions in the numerator and add and subtract the terms in red (as was done in the Product Rule). In the fourth step we inverted h (in the denominator) and multiplied it to simplify the compound fraction.   Noting as before, that the limit of f(x+h) as h approaches 0 is f(x) and likewise the limit of g(x+h) as h approaches 0 is g(x) and that f(x) and g(x) are independent of h and are therefore unaffected by h as h approaches 0 we can write these terms as the following.     Note the “ ― ” sign follows f(x) as we reorder the terms to have first. #8 Chain Rule Given the functions u(x) and g(u(x)) such that u'(x0) and g'(u(x0)) exist and let F(x) name g(u(x)) that is let F(x) = g(u(x)). The following demonstration shows the validity of the chain rule and also the main idea underlying the proof that follows. It will become convenient to used the prime notation for derivatives, that is, d/dx f(x) = f '(x). There may be some x where u(x+h) – u(x) equals zero. What we need to do is eliminate those cases where u(x+h) – u(x) can be zero for some x and non-zero h. We do this by defining a secondary or auxiliary function as we will see. The chain rule theorem is: If g is differentiable at x and f is differentiable at g(x), then their composition f ◦ g is differentiable at x and (f ◦ g )' = f'(g(x)) g'(x) (Recall, f ◦ g is f(g).) The requirements of this theorem are: g is differentiable at x, that is, g is also continuous at x; f is differentiable at g(x), that is, f is also continuous at g(x). This proof takes a few turns that need explaining. Our objective is to prove the Chain Rule without the zero issue above. First of will we need the following form for the limit for the derivative of f at x. We assign G(t) naming the difference quotient for convenience. Equation #1: We will need this equation soon. Then f(t) is differentiable at x iff The theorem states that f(x) is differentiable at x, so this limit must exist. Say this limit equals L, then it must also be the case that Remember, h = t – x, so G(x + h) = G(x + t – x) = G(t). To prove this theorem we use Equation 1 and show that: Equation 2: Note: we have 't' in the denominator, not g(t). Both functions use the same limit variables and bounds since their derivatives are being evaluated at the same x with the same t approaching x. Remember G is the difference quotient for f. First we need to define an auxiliary function. This is a common technique that allows the proof to proceed in a simpler fashion. At first this may seem odd but it will make sense as you learn more proofs. The auxiliary function is F(y). First note the variable change, the independent variable for F is y, not x. Next, to guard against division by zero (see above) we define F(y) in two parts: when y ≠ g(x): Since y ≠ g(x), y – g(x) ≠ 0. when y = g(x): F(y) = f' (g(x)) f'(g(x)) is part of our result and is the limit as y → g(x). Now: The right hand side of this equation, by definition, is the derivative of f at g(x), that is f'(g(x)). This is also the value at F(g(x)) (when y = g(x)). For t ≠ x We start with the left side of equation 2 above, and write it in terms of two difference quotients. So do this we multiply by the ratio in red which equals 1. Equation 3: In step two we interchange the denominators on the right hand side and end up with two difference quotients. The first difference quotient we recognize as F(g(t)). If g(t) = g(x) then both sides equal zero, (because: f(g(t)) then equals f(g(x)), so f(g(t))-f(g(x)) = 0.) and F(g(t)) at g(t) = g(x) was defined earlier to be f'(g(x)). If g(t) ≠ g(x) then we can invert and multiply the difference quotient on the right hand side since we will not be dividing by zero and solve for F(g(t)) like so: The theorem states that g is differentiable at x, so g is also continuous at x. We've shown that F is continuous at g(x), and our study of limits and continuity tells us that this means the composition F ◦ g is also continuous at x. Which means the following limit is valid. Using the definition of F(x) above): This last equation states that (f ◦ g)' = f'(g) g', that is, (f(g(x))' = f'(g(x)) g'(x). This is another FREE TEMPLATE PRINTABLE presented to you from the Template section of K12math.com

Download our free math lesson plan template...and print!!