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Calculus 1  Problems & Solutions  –  Chapter 2  –  Section 2.3.1

 

2.3.1
Differentiation Of Sums, Differences, And
Polynomials

 

 

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Review

 

1. Differentiation Of Sums And Differences

 

 

Theorem 1.1

 

Let f and g be differentiable at x. Then f + g and f – g are differentiable at x, and:

( f + g)'(x) = f '(x) + g'(x),
( f – g)'(x) = f '(x) – g'(x).

 

 

Proof

 

The proof that ( f – g)'(x) = f '(x) – g'(x) is similar.
EOP

 

In words, the derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the
derivatives.

 

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2. Differentiation Of Constant Functions

 

Recall that a constant function is one that is, well, constant, ie, one whose values at all points in its domain are the same.
The graph of such a function is a horizontal line, as shown in Fig. 2.1. Since a constant function doesn't change, its rate
of change, or derivative, is 0.

 

Fig. 2.1

 

A constant function.

 

Theorem 2.1

 

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

 

 

Proof

For any point x we have:

 


EOP

 

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3. Differentiation Of Constant Multiples Of Functions

 

Let f(x) be a function and c a constant. The function cf defined by (cf )(x) = c.f(x) is a constant multiple of f(x). Note
that a constant multiple of f(x) isn't a constant function unless f(x) is. We'll see in the following theorem that the
derivative of c.f(x) equals c multiplied by the derivative of f(x).

 

Theorem 3.1

 

Let f be differentiable at x and c a constant. Then cf is also differentiable at x, and:

 

(cf )'(x) = c . f '(x).

 

 

Proof

EOP

 

In words, the derivative of a constant multiple is the constant multiple of the derivative.

 

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4. Differentiation Of Positive Integer Powers Of x

 

The proof of the following theorem uses the factorization of the difference of nth powers:

 

an – bn = (a – b) (an–1 + an–2b + an–3b2 + ... + a2bn–3 + abn–2 + bn–1),

 

which can be checked by multiplying out the right-hand side.

 

Theorem 4.1

 

Let n be a positive integer. Then:

 

 

 

Proof

Using the factorization of the difference of nth powers we get:

 

EOP

 

It may be easier to memorize the formula in this theorem in this form: (xn)' = nxn–1.

 

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5. Differentiation Of Polynomials

 

Theorems 1-4 show that the derivative of the polynomial:

 

P(x) = a0 + a1x + a2x2 + a3x3 + ... + an–1xn–1 + anxn

 

is:

 

P '(x) = a1 + 2a2x + 3a3x2 + ... + (n – 1)an–1xn–2 + nanxn–1.

 

 

Why isn't there the factor x in the “zeroth” term a0 of P(x)? Well, there is, but its exponent is 0, like the subscript of the
factor a0, and x0 = 1, so that a0x0 = (a0)(1) = a0. What's the exponent of x in the “first” term a1x? Well, it’s 1, like the
subscript of a1; remember, x1 = x. So the polynomial P(x) can be written in concise form as follows:

 

 

Example 5.1

 

Find the derivative of y = x4 – 2x3 + 3x2 – 4x + 5.

 

Solution
 y' = 4x3 – 6x2 + 6x – 4.
EOS

 

Example 5.2

 

Calculate f '(2) if f(x) = 3x2 – 5x.

 

Solution
We have f '(x) = 6x – 5. Hence f '(2) = 6(2) – 5 = 7.

EOS

 

 f '(2) is the derivative of f(x) at x = 2, or the value of f '(x) at x = 2, not the derivative of f(2) (the derivative of f(2) is 0
because f(2) is a constant). That's why we calculate f '(x) first, then we substitute x = 2 in f '(x).

 

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Problems & Solutions

 

1.  Differentiate f(x) = (2x – 5)(3 – 6x).

 

Solution

 

 f(x) = 6x – 12x2 – 15 + 30x = – 12x2 + 36x – 15,
 f '(x) = – 24x + 36.

 

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2.  Find (d/dt) g(t) if:

 

   

 

Solution

 

 

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3.  Let f(x) = x(3x– 5)2 and I = [–1, 3].
     a.  Find the average rate of change of f over I.
     b.  Find the instantaneous rate of change of f at the midpoint of I.

 

Solution

 

     Now:

 

      f(x) = x(3x – 5)2 = x(9x2 – 30x + 25) = 9x3 – 30x2 + 25x.

 

     So f '(x) = 27x2 – 60x + 25. Thus, the instantaneous rate of change of f at the midpoint of I is:

 

     f '(1) = 27(1)2 – 60(1) + 25 = – 8.

 

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4.  Let C be the curve with equation y = 4x2 – x4.
     a.  Find all the horizontal tangent lines to C.
     b.  Find the normal line to C at x = 1/2.

 

Solution

 

a.

 

 

Note

 

Recall that the notation y(a) means the value of y at x = a, not the product ya.

 

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5.  The equation xy' = 3y involves the derivative of a function and thus is called a differential equation.
     a.  Show that for any constant C, the function y = Cx3 satisfies the given equation. That function is called a solution of
          the given differential equation.
     b.  Determine the particular solution y = f(x) of the given differential equation that passes thru the point (2, 5).

 

Solution

 

a.  From y = Cx3 we have y' = 3Cx2. So xy' = x(3Cx2) = 3(Cx3) = 3y, which shows that y = Cx3 satisfies the equation
     xy' = 3y.

 

b.  5 = f(2) = C(2)3 = 8C, thus C = 5/8. Consequently y = (5/8)x3 is the particular solution that passes thru the point (2,
     5).

 

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6.  Suppose that when a pebble is dropped into a tank of water, a wave travels outward in a circular ring whose radius
     increases at a constant rate of 20 cm/sec.
     a.  Find the area of the circular disk enclosed by the wave 3 seconds after the pebble hits the water.
     b.  Find the instantaneous rate at which the area of the disk is increasing 3 seconds after the pebble hits the water.

 

Solution

 

 

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