Guide - Techniques to Finding Derivatives Product Rule Given: A function f (x) that is the product of two functions u(x) and v(x). f (x) = u(x)v(x)

Examples: f (x) = x2 (x + 5)2 g(x) = (x + 7)3 ex √ r(x) = ( x + 7)(x3 + 9) h(x) = (3x − 9)2

Rule: First, identify your u(x) and v(x). Then, f 0 (x) = u(x)v 0 (x) + v(x)u0 (x) Sometimes read as “First times derivative of second plus second times derivative of first.” Worked Example:

√ f (x) = 4x3 ( x + 7)

Then we have: u(x) = 4x3 u0 (x) =√ 12x2 1 v(x) = x + 7 = x 2 + 7 −1 v 0 (x) = 12 x 2 Plugging into the product rule yields: f 0 (x) = u(x)v 0 (x) + v(x)u0 (x) 1 1 −1 f 0 (x) = (4x3 )( x 2 ) + (x 2 + 7)(12x2 ) 2 5 5 = 2x 2 + 12x 2 + 12x2 5 = 14x 2 + 12x2

Note: Due to the fact that addition is commutative, it doesn’t matter if you write the product rule as uv 0 + vu0 or as vu0 + uv 0 , they give the same result. The advantage to thinking the latter way in mentioned after the Quotient Rule.

Quotient Rule Given: A rational function f (x) =

u(x) v(x)

Examples: f (x) =

6x + 1 3x + 10

x2 − 4x 4x2 − 5 √ x+3 y= 100 − x2

g(x) =

Rule: First note that u(x) is the numerator and v(x) is the denominator. Then: v(x)u0 (x) − u(x)v 0 (x) [v(x)]2 Sometimes read as “Bottom times derivative of top minus top times derivative of bottom all over bottom squared” f 0 (x) =

Worked Examples: Ex. g(x) =

x2 + 3 4 − 5x

Identifying u(x), u0 (x), v(x) and v 0 (x) gives: u(x) = x2 + 3 u0 (x) = 2x v(x) = 4 − 5x v 0 (x) = −5 Then just plug it all in: vu0 − uv 0 v2 (4 − 5x)(2x) − (x2 + 3)(−5) = (4 − 5x)2 (8x − 10x2 ) − (−5x2 − 15) = (4 − 5x)2 −5x2 + 8x + 30 = 25x2 − 40x + 16

g 0 (x) =

Ex. f (x) =

4x2 − 3 √ 3 x

Identifying u(x), u0 (x), v(x) and v 0 (x) gives: u(x) = 4x2 − 3 u0 (x) =√ 8x 1 3 v(x) = x = x 3 −2 v 0 (x) = 13 x 3 Plugging this all in yields:

f 0 (x) = = =

vu0 − uv 0 v2 −2 1 (x 3 )(8x) − (4x2 − 3)( 31 x 3 ) 1

[x 3 ]2 −2 4 4 8x 3 − 43 x 3 + 3x 3 2

x3 2

−4 20x 3 = + 3x 3 3

A way to remember: Most people can remember the product rule but some fumble on the quotient rule. Here’s a way to remember it: • First write: “vu0 space uv 0 ” • If you want the product rule, just throw a “+” in the space. • If you want the quotient rule, throw a “–” in the space and put it over [v(x)]2 • Order is important, this trick will not work if you write “uv 0 is not commutative.

vu0 ” since subtraction

Chain Rule Given: A composition of functions u(x) and v(x). f (x) = v[u(x)] Note: This is most often seen as an expression in x raised to a power. Examples: y = (x3 + 1)70 where u(x) = x3 + 1 and v(x) = x7 0 √ √ 4 f (x) = x2 − 1 where u(x) = x2 − 1 and v(x) = 4 x √ √ y = ( x − 5)3 where u(x) = x − 5 and v(x) = x3 g(x) = (x − a)b where u(x) = x − a and v(x) = xb Rule: If we have f (x) = v[u(x)], then: f 0 (x) = v 0 [u(x)] ∗ u0 (x) (Intuitively:) We are given something of the form, for example, f (x) = 2(x3 + 5x)6 Ignore the x3 + 5x at first, leave it alone. Just bring the 6 down like in the power rule, and subtract one from the exponent, just like in the power rule to get: 12(x3 + 5x)5

(1)

Now, the chain rule’s actual rule says we have to multiply the expression in line (1) by the derivative of the stuff we ignored. The derivative of x3 + 5x is, as we know, 3x2 + 5. So the derivative is: f 0 (x) = 12(x3 + 5x)5 (3x2 + 5) = (36x3 + 60)(x3 + 5x)5 and that’s the answer. (Formally:) I mentioned once that the notation f 0 (x) and dy/dx are equivalent. We read dy/dx as “The derivative of y with respect to x”. This means that we have taken the derivative of a function, y, that has variable x. For example, y = x5 f 0 (x) =

dy = 5x4 dx

s = t2 − 5t ds = 2t − 5 dt

In the second example, we have a function s in terms of variable t. So we could say s0 (t) or ds/dt. Get comfortable with that before reading on. Good? Now again we are given y = 2(x3 + 5x)6 and we want to find dy/dx. This time, let’s not ignore the x3 + 5x and instead just set it equal to u. So u = (x3 + 5x) and we have y = 2u6 . Now we have: y = 2u6 dy = 12u5 du However, we want dy/dx, not dy/du. Now compute du/dx, the derivative of u with respect to x. We get: u = x3 + 5x du = 3x2 + 5 dx Now notice that we can say: dy dy du = ∗ , dx du dx since the du terms will cancel. Finally we have: dy du dy = ∗ = 12u5 ∗ (3x2 + 5) = 12(x3 + 5x)5 (3x2 + 5) = 36x3 + 60(x3 + 5x) dx du dx Which is the same answer as above, as desired. Worked examples: g(x) = 3(2x − 7)4 g 0 (x) = 12(2x − 7)3 (2) = 24(2x − 7)3 (2x2 − 7)3 1 = (2x2 − 7)3 12 12 2 2 3(2x − 7) )(4x) f 0 (x) = = x(2x2 − 7)2 = 4x5 − 28x3 + 49x 12

f (x) =

Power Rule Rule: Given f (x) = axn . We have f 0 (x) = naxn−1 You may ask, “Lucas, why are you putting the Power Rule fourth on the list instead of first?” Excellent question, curious student. Well, when you think about it, the Power Rule is just a special case of the Chain Rule, so it makes sense for it to come after. If you consider the “stuff inside parentheses” in the Chain Rule to be just x, you get the derivative you multiply by to be 1. Thus, the Power Rule really does follow immediately from the Chain Rule.

Exponential Rule:

Preliminary Fact: The function ex is very special, more than you realize. It happens to be one of the very few functions that is its own derivative. That is, if you have: f (x) = ex Then f 0 (x) = ex

Given: A function of the form: f (x) = au(x) for some function u(x). Examples: f (x) = e2x 2 f (x) = e3x −2 3 y = 10x −5x Rule: The derivative is f 0 (x) = ln(a) ∗ au(x) ∗ u0 (x) Note: I will almost always give you problems where the value of a is e. This gives the specific rule: f (x) = eu(x) f 0 (x) = eu(x) u0 (x) This result is immediate from using the Chain Rule and Preliminary Fact. Worked Example: Let f (x) = 5ex

2 −5

Then it’s immediate that we have 2 −5

f 0 (x) = 5ex

(2x) = 10xex

2 −5

Logarithm Rule:

Given: A function in the form f (x) = loga [u(x)] Examples: r(x) = 3 log2 x f (x) = 5 log(7x) g(x) = loge (2x) = ln(2x) Rule: The derivative of y = k loga [u(x)] for some numbers k and a and some function u(x) is dy ku0 (x) = dx ln(a)u(x) Note: I will almost always give you exercises where the value of a is e, this gives the specific formula where f (x) is in the form f (x) = k ln[u(x)] and then the derivative is f 0 (x) =

ku0 (x) u(x)

Worked Examples: Ex. Let f (x) = ln(2x). Find f 0 (x). The rule says: f 0 (x) =

1∗2 1 = 2x x

Ex. Let g(x) = 10 ln(x2 + 5x). Find the derivative. The rule says: g 0 (x) =

10(2x + 5) 20x + 50 = 2 2 x + 5x x + 5x

Ex. Let y = 5 log2 (x3 ). Then we have: dy 5(3x2 ) 15x2 = = dx ln 2(x3 ) ln 2(x3 )

Practice Problems Warm-up 1. x2 2. (x + 2)(x − 3) 3. e2x 4. x3 +

x 5

5. xex

Find the derivative. 1. 4(x + 5)3 2. 4x(x + 5)3 3. 4e2x 4. 4xe2x 5.

x+3 x−1

6. (2x3 + 1)3 (x − 2) 7. 4 ln(x + 3)

8. 4x ln(x + 3) x2 − 5 (x − 5)2 8 ln(x) 10. y = x2 x 4 x2 e 11. f (x) = 24 9.

√

12. g(x) = ln(xe

x

+ 4)

√ 13. (x − 3) 3 x + 4 14. 10x 15.

2

(x + 4)2 (x − 3)2 ex