Calculus: The Language of Change

Exercise Set 6.1
Rules for Special Functions

by Sharon Portz

Exercise 1 : Power Rule does not apply here

You cannot divide by zero or take even roots of negative numbers (as real functions).  Show that  the Power Rule does not apply at x = 0 for p = (1/3) or at x = -2 for p = (1/4).

y = (x^(p ))= (x^((1/3)))   =>  (dy/dx) = p (x^(p - 1)) = ?
		
	and
	
	y = (x^(p ))= (x^((1/4))) => (dy/dx) = p (x^(p - 1)) = ?

Use the computer to graph the two functions y=(x^((1/3))) = (x^(1/3)) and y=(x^((1/4))) = (x^(1/4)) for -3 ≤ x ≤ 3.  Explain your "bad" analytical result above in terms of the graph.

y = (x^(p ))= (x^((1/3)))

(dy/dx) = p (x^(p - 1)) = ?

Solution to y = (x^((1/3)))

(dy/dx) = (1/3)(x^((1/3) - 1)) = (1/3)(x^(((-2)/3))) = (1/(3 (x^(2/3))))

When x = 0 we have
(dy/dx) =(1/(3 (0^(2/3)))) = (1/0) which is undefined.
Therefore the Power Rule does not apply when x = 0.

y = (x^(p ))= (x^((1/4)))

(dy/dx) = p (x^(p - 1)) = ?

Solution to y = (x^((1/4)))

(dy/dx) = (1/4)(x^((1/4) - 1)) = (1/4)(x^(((-3)/4))) = (1/(4 (x^(3/4))))

When x = -2 we have
(dy/dx) =(1/(4 (((-2))^(3/4))))  which is undefined since the even root of a negative number is not a real number.
Therefore the Power Rule does not apply when x = -2.

computer graph

Solution to computer graph and explanation

Graphing the two functions y=(x^((1/3))) = (x^(1/3)) and y=(x^((1/4))) = (x^(1/4)) looks like this:

We see that the slope is vertical (or undefined) at 0.

Exercise 2 : Power Rule

a) Show that the Power Rule agrees with all the derivatives we computed directly in and before Exercise 5.1.1 as well as those that you computed in Problem 5.1.  They are the following:

y = x => (dy/dx) = 1 = 1 (x^(1 - 1)) = (x^0)			y = (1/x) = (x^(-1)) => (dy/dx) = ((-1)/(x^2)) = -1 (x^(-2))

y = (x^2) => (dy/dx) = 2 (x^1) = 2 (x^(2 - 1))			y = (1/(x^2)) = (x^(-2)) => (dy/dx) = -(2/(x^3)) = -2 (x^(-3))

y = (x^3) =>(dy/dx) = 3 (x^2) = 3 (x^(3 - 1))			y = (Sqrt[x]) = (x^((1/2))) => (dy/dx) = (1/(2 (Sqrt[ x]))) = (1/2)(x^(-(1/( 2))))

y = (x^n) => (dy/dx) = n (x^(n - 1))				y = (x^(1/3)) = (x^((1/3))) => (dy/dx) = (1/(3 (( (x^2))^(1/3)))) = (1/3)(x^(-(2/3)))

b) Differentiate the following by first converting to power form and then applying the Power Rule.  Convert your derivatives back to radical notation.

a) y = (1/(x^5))			b) y = (x^(1/5))			c) y = \(TraditionalForm\`1\/\@\(x\^2\)\%3\)

d) y = (x^2) (Sqrt[x])			e) y = (x^4) (Sqrt[(x^3)])			f) y = ((x^2)/(Sqrt[x]))

g) y = x (x^(1/3))			h) y = (x^5)(((x^2))^(1/3))			i) y = \(TraditionalForm\`x\^2\/\@\(x\^2\)\%3\)

c) The derivative of a constant funcion is zero.  Why? The Power Rule also includes a case of this in the form,

y = (x^0) => (dy/dx) = 0 = 0 (x^(0 - 1))

What is the value of(x^0) ?  Is (0^0) defined?

Part a

a) Show that the Power Rule agrees with all the derivatives we computed directly in and before Exercise 5.1.1 as well as those that you computed in Problem 5.1.  They are the following:

y = x => (dy/dx) = 1 = 1 (x^(1 - 1)) = (x^0)			y = (1/x) = (x^(-1)) => (dy/dx) = ((-1)/(x^2)) = -1 (x^(-2))

y = (x^2) => (dy/dx) = 2 (x^1) = 2 (x^(2 - 1))			y = (1/(x^2)) = (x^(-2)) => (dy/dx) = -(2/(x^3)) = -2 (x^(-3))

y = (x^3) =>(dy/dx) = 3 (x^2) = 3 (x^(3 - 1))			y = (Sqrt[x]) = (x^((1/2))) => (dy/dx) = (1/(2 (Sqrt[ x]))) = (1/2)(x^(-(1/( 2))))

y = (x^n) => (dy/dx) = n (x^(n - 1))				y = (x^(1/3)) = (x^((1/3))) => (dy/dx) = (1/(3 (( (x^2))^(1/3)))) = (1/3)(x^(-(2/3)))

Solution to part a

Here all you need to do is look at the equations above which you have already computed.  Notice how the Power Rule was used to compute the derivatives.  Recall that the Power Rule is

y = (x^n) => (dy/dx) = n (x^(n - 1))

The values for n in the above equations are as follows:
 n = 1						n = -1

n = 2 						n = -2

n = 3						n = (1/2)

n = n						n = (1/3)

Part b

b) Differentiate the following by first converting to power form and then applying the Power Rule.  Convert your derivatives back to radical notation.

a) y = (1/(x^5))			b) y = (x^(1/5))			c) y = \(TraditionalForm\`1\/\@\(x\^2\)\%3\)

d) y = (x^2) (Sqrt[x])			e) y = (x^4) (Sqrt[(x^3)])			f) y = ((x^2)/(Sqrt[x]))

g) y = x (x^(1/3))			h) y = (x^5)(((x^2))^(1/3))			i) y = \(TraditionalForm\`x\^2\/\@\(x\^2\)\%3\)

Solution to part b

First we need to find the value of n, the power of x.  See below for each solution which was found using WalkD.

y = (1/(x^5))

Here, n = -5

y = (x^(1/5))

n  =  (1/5)

y = \(TraditionalForm\`1\/\@\(x\^2\)\%3\)

\(TextForm
    \`Here\ we\ have\ n\  = 
      \(-\ \(2\/3\) . 
			\ \ Notice\)\ in\ the\ following\ solution\ by\ 
        WalkD\ the\ computer\ used\ the\ Linearity\ Rule\ and\ the\ Power\ 
        rule . \ \ Don' t\ mind\ the\ Null\ term\ in\ the\ expressions\ 
        below . \ \ It\ comes\ from\ having\ the\ 1\ in\ the\ 
        \(numerator . \)\)

y = (x^2) (Sqrt[x])

n = 2 + (1/2) = (5/2)

y = (x^4) (Sqrt[(x^3)])

n  =  4  +  (3/2)  =  (11/2)

y = ((x^2)/(Sqrt[x]))

n = 2 - (1/2) = (3/2)

y = x (x^(1/3))

n = 1 + (1/3) = (4/3)

y = (x^5)(((x^2))^(1/3))

n = 5 + (2/3) = (17/3)

y = \(TraditionalForm\`x\^2\/\@\(x\^2\)\%3\)

n = 2 - (2/3) = (4/3)

Part c

c) The derivative of a constant function is zero.  Why? The Power Rule also includes a case of this in the form,

y = (x^0) => (dy/dx) = 0 = 0 (x^(0 - 1))

What is the value of (x^0) ?  Is (0^0) defined?

Solution to part c

The derivative of a constant function is zero because of the Power Rule.  Let's say our constant function is f[x] = 6.  That is the same as f[x] = 6 (x^0) because (x^0) = 1.  Using the Power Rule we get:
0(6)((x^(0 - 1))) = 0.  This is true for all constant functions.  By the way, (0^0) is undefined by convention.

Exercise 3: Other Variables

a)  y = (x^2) => (dy/dx) = ?
b) u = (1/(v^2)) =>(du/dv) = ?

c) y = (Sqrt[x]) => (dy/dx) = ?
d) u = (1/(Sqrt[v])) =>(du/dv) = ?

e) y = (x^3)(Sqrt[(x^3)]) => (dy/dx) = ?
f) u = \(TraditionalForm\`v\^2\/\@\(v\^2\)\%3\)  =>(du/dv) = ?

g) y = Sin[x] => (dy/dx) = ?
h) u = Cos[v] =>(du/dv) = ?

i) y = Log[x] => (dy/dx) = ?
j) u = (e^v) =>(du/dv) = ?

Part a

a)  y = (x^2) => (dy/dx) = ?

Solution to a

Part b

b) u = (1/(v^2)) =>(du/dv) = ?

Solution to b

Part c

c) y = (Sqrt[x]) => (dy/dx) = ?

Solution to c

Part d

d) u = (1/(Sqrt[v])) =>(du/dv) = ?

Solution to d

Part e

e) y = (x^3)(Sqrt[(x^3)]) => (dy/dx) = ?

Solution to e

Part f

f) u = \(TraditionalForm\`v\^2\/\@\(v\^2\)\%3\)  =>(du/dv) = ?

Solution to f

Part g

g) y = Sin[x] => (dy/dx) = ?

Solution to g

Part h

h) u = Cos[v] =>(du/dv) = ?

Solution to h

Part i

i) y = Log[x] => (dy/dx) = ?

Solution  to i

Part j

j) u = (e^v) =>(du/dv) = ?

Solution to j