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# 8.5: Techniques of Differentiation

Created by: CK-12

## Learning objectives

• Use various techniques of differentiations to find the derivatives of various functions.
• Compute derivatives of higher orders.

In the last lesson we calculated derivatives using the limit definition. In this lesson, we will develop formulas and theorems that will calculate derivatives in more efficient and quick ways. Look for these theorems in boxes throughout the lesson.

## Constants

Theorem: If f(x) = c where c is a constant, then f '(x) = 0.

Proof: $f'(x)= \lim_{h \to 0}\frac {f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{c-c}{h} = 0$

Example 1: Find f '(x) for f(x)=16.

Solution: If f(x) = 16 for all x, then f '(x) = 0 for all x. We can also write $\frac{d}{dx}16 = 0.$

Theorem: If c is a constant and f is differentiable at all x, then $\frac {d}{dx}[cf(x)] = c\frac {d}{dx}[f(x)]$. In simpler notation (cf)' = c(f)' = cf '.

Example 2: Find the derivative of f(x)=4x3.

Solution: $\frac {d}{dx}\left [{4x^3} \right ]= 4 \frac{d}{dx}\left [{x^3} \right ] = 4\left [{3x^2} \right ] = 12x^2$

Example 3: Find the derivative of $f(x)=\frac{-2}{x^{4}}$.

Solution: $\frac {d}{dx} \left [ \frac{-2}{x^4} \right ]=\frac {d}{dx} \left [{-2x^{-4}} \right ]= -2\frac {d}{dx} \left [{x^{-4}} \right ] = -2 \left [{-4x^{-4-1}} \right ] = -2 \left [{-4x^{-5}} \right ] = 8x^{-5} =\frac {8}{x^5}$

## The Power Rule

Theorem: (The Power Rule) If n is a positive integer, then for all real values of x

$\frac {d}{dx}[x^n] = nx^{n-1}$.

Example 4: Find the derivatives of:

1. $f(x)=x^{3}$
2. $f(x)=x$
3. $f(x)=\sqrt{x}$
4. $f(x)=\frac{1}{x^{3}}$

Solution:

1. If f(x)=x3 then f'(x) = 3x3-1 = 3x2
2. $\frac {d}{dx}[x] = 1x^{1-1} = x^0 = 1$
3. $\frac {d}{dx}[\sqrt{x}] = \frac {d}{dx}[x^{1/2}] =\frac {1}{2}x^{1/2-1} =\frac {1}{2}x^{-1/2} =\frac{1}{2x^{1/2}} =\frac {1}{2\sqrt{x}}$
4. $\frac {d}{dx}\left [ \frac{1}{x^3} \right ]= \frac {d}{dx}\left [{x^{-3}} \right ]= -3x^{-3-1} = -3x^{-4} = \frac {-3}{x^4}$

## Derivatives of Sums and Differences

Theorem: If f and g are two differentiable functions at then

$\frac {d}{dx}\left [{f(x)+g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]+\frac {d}{dx}\left [{g(x)} \right ]$

and

$\frac {d}{dx}\left [{f(x)-g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]- \frac {d}{dx}\left [{g(x)} \right ]$.

In simpler notation

$(f \pm g)'= f'\pm g'$.

Example 5: Find the derivatives of:

1. $f(x)=3x^2+2x$
2. $f(x)=x^3-5x^2$

Solution:

1. Notice that we use the power rule to help:
$\frac {d}{dx}\left [{3x^2+2x} \right ]$ $= \frac {d}{dx}\left [{3x^2} \right ]+\frac {d}{dx}\left [{2x} \right ]$
$= 3\frac {d}{dx}\left [{x^2} \right ]+2 \frac {d}{dx}\left [{x} \right ]$
$= 3 \left [{2x} \right ]+2 \left [{1} \right ]$
$= 6x+2$
2. Again, we use the power rule to help:
$\frac {d}{dx}\left [{x^3-5x^2} \right ]$ $= \frac {d}{dx}\left [{x^3} \right ]-5 \frac {d}{dx}\left [{x^2} \right ]$
$= 3x^2- 5 \left [{2x} \right ]$
$= 3x^2-10x$

## The Product Rule

Theorem: (The Product Rule) If f and g are differentiable at x, then$\frac {d}{dx}\left [{f(x)\cdot g(x)} \right ]= f(x) \frac {d}{dx}g(x) + g(x)\frac {d}{dx}f(x)$

In a simpler notation

$(f\cdot g)'=f\cdot g'+g\cdot f'$

In words, the derivative of the product of two functions is equal to the first times the derivative of the second plus the second times the derivatives of the derivative of the first.Keep in mind that

$(f\cdot g)' \neq f'+g'$

Example 6: Find (dy/dx) for y = (3x4 + 2)(7x3 - 1).

Solution: There two methods to solve this problem. One is to multiply the product and then use the derivative of the sum rule. The second is to directly use the product rule. Either rule will produce the same answer. We begin with the sum rule.

y = (3 x 4 + 2)(7 x 3 -1)
= 21 x 7 -3 x 4 + 14 x 3 -2
Taking the derivative of the sum yields
$\frac {dy}{dx}$ $= {147x^6-12x^3+42^2+0}$
$= 147x^6-12x^3+42x^2$
Now we use the product rule.
$y'$ $= (3x^4+2)\cdot (7x^3-1)'+(3x^4+2)'\cdot (7x^3-1)$
$= (3x^4+2)(21x^2)+(12x^3)(7x^3-1)$
$= (63x^6+42x^2)+(84x^6-12x^3)$
$= 147x^6-12x^3+42x^2$

## The Quotient Rule

Theorem: (The Quotient Rule) If f and g are differentiable functions at x and g(x) ≠ 0, then$\frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}$

In simpler notation

$\left ( \frac{f}{g} \right )'=\frac {g\cdot f'-f\cdot g'}{g^2}$

Keep in mind that the order of operations is important (because of the minus sign in the numerator) and $\left ( \frac{f}{g} \right )'\neq \frac {f'}{g'}$.

Example 7: Find dy/dx for $y = \frac {x^2-5}{x^3+2}$

Solution:

$\frac {dy}{dx}$ $= \frac {d}{dx}\left [ \frac{x^2-5}{x^3+2} \right ]$
$= \frac {(x^3+2).(x^2-5)'-(x^2-5).(x^3+2)'}{(x^3+2)^2}$
$= \frac {(x^3+2)(2x)-(x^2-5)(3x^2)}{(x^3+2)^2}$
$= \frac {2x^4+4x-3x^4+15x^2}{(x^3+2)^2}$
$= \frac {-x^4+15x^2+4x}{(x^3+2)^2}$
$= \frac {x(-x^3+15x+4)}{(x^3+2)^2}$

Example 8: At which point(s) does the graph of $y = \frac {x} {x^2+9}$ have a horizontal tangent line?

Solution: Since the slope of a horizontal line is zero, and since the derivative of a function signifies the slope of the tangent line, then taking the derivative and equating it to zero will enable us to find the points at which the slope of the tangent line equal to zero, i.e., the locations of the horizontal tangents. Notice that we will need to use the quotient rule here:

$y$ $= \frac{x} {x^2 + 9}$
$y'$ $= \frac{(x^2 + 9) (1) - x (2x)} {(x^2 + 9)^2} = 0$
Cross-multiplying,
$x^2 + 9 - 2x^2$ $= 0$
$x^2$ $= 9$
$x$ $= \pm 3$
Therefore, at x = -3 and +3, the tangent line is horizontal.

## Higher Derivatives

If the derivative $\,\! f'$ of the function $\,\! f$ is differentiable, then the derivative of $f'$, denoted by $\,\! f''$ is called the second derivative of $\,\! f$. We can continue the process of differentiating derivatives and obtain third, fourth, fifth and higher derivatives of $\,\! f$. They are denoted by $\,\! f'$, $\,\! f''$, $\,\! f'''$, $\,\! f^{(4)}$, $\,\! f^{(5)}$, . . . ,

Example 9: Find the fifth derivative of f(x) = 2x4 - 3x3 + 5x2 - x - 1.

Solution: To find the fifth derivative, we must first find the first, second, third, and fourth derivatives.

f '(x) = 8x3 - 9x2 + 5x - x
f ' (x) = 24x2 - 18x + 5
f "' (x) = 48 x - 18
f(4) (x) = 48
f(5) (x) = 0

## Lesson Summary

In this lesson we moved from finding the derivative of functions using the formal definition of a derivative to finding the derivative of functions using theorems. These theorems are like shortcuts that allow us to quickly find derivatives of functions. While we did not prove most of the theorems in this lesson, when you study calculus you will become familiar with these proofs.

## Points to Consider

1. If the first derivative can be interpreted as instantaneous rate of change, e.g. the instantaneous velocity of an object, what can the second derivative be interpreted as?
2. Can you prove the product rule using other theorems in this lesson?

## Review Questions

Use the results of this section to find the derivatives (dy/dx) of the following:

1. y = 5x7
2. y = -3x
3. $f(x) = \frac{1} {3} x + \frac{4} {3}$
4. $y = x^4 - 2x^3 - 5\sqrt{x} + 10$
5. y = (5x2 - 3)2
6. $y = \frac{1} {2} (x^3 - 2x^2 + 1)$
7. $y = \sqrt{2} x^3 -\frac{1} {\sqrt{2}}x^2 + 2x + \sqrt{2}$
8. y = a2 - b2 + x2 - a - b + x (where a, b constants)
9. $y = x^{-3} + \frac{1} {x^7}$
10. y = (x3 - 3x2 + x)(2x3 + 7x4)
11. $y = \left (\frac{1} {x} + \frac{1} {x^2}\right ) (3x^4 - 7)$
12. $y = \sqrt{x} + \frac{1} {\sqrt{x}}$
13. $y = \frac{3} {\sqrt{x} + 3}$
14. $y = \frac{4x + 1} {x^2 - 9}$
15. Newton’s Law of Universal Gravitation states that the gravitational force between two masses (say, the earth and the moon), m and M is equal two their product divided by the squared of the distance r between them. Mathematically, $F = G\frac{mM} {r^2}$ where G is the Universal Gravitational Constant (1.602 × 10-11 Nm2/kg2). If the distance r between the two masses is changing, find a formula for the instantaneous rate of change of F with respect to the separation distance r.
16. Find $\frac{d} {d \psi} \left [\frac{\psi \psi_0 + \psi^3} {3 - \psi_0}\right ]$ where $\psi_0$ is a constant.
17. Find $\frac{d^3y} {dx^3} |_{x = 1},$ where $y = \frac{2} {x^3}$.

1. y ' = 35x6
2. -3
3. $\frac{1} {3}$
4. $4x^3 - 6x^2 - \frac{5} {2 \sqrt{x}}$
5. 20x(5x2 - 3)
6. $y' = \frac{3} {2} x^2 - 2x$
7. $y' = 3\sqrt{2} x^2 - {\sqrt{2}} x + 2$
8. y' = 2x + 1
9. $y' = \frac{-3} {x^4} - \frac{7} {x^8}$
10. y ' = (x3 - 3x2 + x)(6x2 + 28x3) + (3x2 - 6x + 1)(2x3 + 7x4)
11. $y' = 9x^2+6x+\frac{7}{x^2}+\frac{14}{x^3}$
12. $y' = \frac{1} {2\sqrt{x}} - \frac{1} {2x^{3/2}}$
13. $y' = \frac{-3} {2\sqrt{x} (\sqrt{x} + 3)^2}$
14. $y' = \frac{-4x^2 - 2x - 36} {(x^2 - 9)^2}$
15. $\frac{dF} {dr} = -2G \frac{mM} {r^3}$
16. $\frac{\psi_0 + 3\psi^2} {3 - \psi_0}$
17. -120

Feb 23, 2012

Jan 13, 2014