CHAPTER 8
The Derivative
8.1 Using the A-definition, find the derivative /'(x) of the function /(x) = 2x - 7.
So,
Hence Thus, Answer
8.2 Using the A-definition, show that the derivative of any linear function /(x) = Ax + B is f ' ( x ) = A.
Then,
Hence. Thus,
8.3 Using the A-definition, find the derivative f ' ( x ) of the function /(x) = 2x2 - 3x + 5.
Thus, Hence,
lim
8.4 Using the A-definition, find the derivative /'(*) of the function f(x) = x3.
So,
So,
Thus,
8.5 State the formula for the derivative of an arbitrary polynomial function f(x) = anx" + an_lx" ' + • • • + a2x2 +
a1x+a0.
8.6 Write the derivative of the function f(x) = lx~ - 3x4 + 6x2 + 3x + 4.
/'(*) = 35x4 - I2x3 + \2x + 3.
8.7 Given functions f(x) and g(x), state the formulas for the derivatives of the sum f(x) + g(x), the product
fix) • e(x), and the quotient f(x) /g(x).
[Notice the various ways of denoting a derivative:
8.8 Using the product rule, find the derivative of f(x) = (Sx3 - 20* + 13)(4;t6 + 2x5 - lx2 + 2x).
F'(x) = (5x3-2Qx+13)(24x5 + Wx4-Ux + 2) + (4x" + 2x5-Ix2 + 2x)(15x2-20). [In such cases, do
not bother to carry out the tedious multiplications, unless a particular problem requires it.]
8.9 Using the formula from Problem 8-7. find the derivative of
49
, 50 CHAPTER 8
8.10 Using the formula from Problem 8.7, find the derivative of
8.11 Using the A-definition, find the derivative of
Hence
So,
and
8.12 Using formulas, find the derivatives of the following functions: (a)
(a) -40x4 + 3V3 x2 + 4Trx. Answer
(b) W2x50 + 36xu - 2Sx + i/7. Answer
8.13 Find the slope-intercept equation of the tangent line to the graph of the function f(x) = 4x3 - 7x2 at the point
corresponding to x = 3.
When x = 3, f(x) - 45. So, the point is (3,45). Recall that the slope of the tangent line is the derivative
/'(*), evaluated for the given value of x. But, /'(*) = 12x2 - Ux. Hence, /'(3) = 12(9) - 14(3) = 66.
Thus, the slope-intercept equation of the tangent line has the form y = 66x + b. Since the point (3,45) is on
the tangent line, 45 = 66(3) + 6, and, therefore, b = -153. Thus, the equation is v=66*-153.
Answer
8.14 At what point(s) of the graph of y = x5 + 4x - 3 does the tangent line to the graph also pass through the point
5(0,1)?
The derivative is y' = 5x4 + 4. Hence, the slope of the tangent line at a point A(xa, y0) of the graph is
5*o + 4. The line AB has slope So, the line AB is the tangent line if and
+4
only if (x0 + 4x0 - 4) Ix0 = 5x1 - Solving, x0 = — 1. So, there is only one point (—1, —8).
8.15 Specify all lines through the point (1, 5) and tangent to the curve y = 3>x3 + x + 4.
y' = 9x2 + l. Hence, the slope of the tangent line at a point (xa, ya) of the curve is 9*0 + 1. The slope of
the line through (x0, y0) and (1,5) is So, the tangent line passes
through (1,5) if and only if = 9x20 + l, 3*2 + j r 0 - l = (je 0 -l)(9*S + l), 3x30 + x0-l=9x30-
9**+ *„-!, 9*0 = 6*0, 6*o-9*o = 0, 3*0(2*0 - 3) = 0. Hence, *0 = 0 or * 0 = | , and the points on
the curve are (0, 4) and (§, ^). The slopes at these points are, respectively, 1 and f . So, the tangent lines are
y — 4 = x and y — *TT = T(X—%), or, equivalently, y = x + 4 and y = S f X — ".
8.16 Find the slope-intercept equation of the normal line to the graph of y = jc3 — x2 at the point where x = l.
The normal line is the line perpendicular to the tangent line. Since y' = 3x2 — 2x, the slope of the tangent
lineal x = 1 is 3(1)2 - 2(1) = 1. Hence, the slope of the normal line is the negative reciprocal of 1, namely
— 1. Thus, the required slope-intercept equation has the form y = —x + b. On the curve, when x = \,
y = (I) 3 - (I) 2 = 0. So, the point (1,0) is on the normal line, and, therefore, 0 = -1 + b. Thus, b = \,
and the required equation is y = — x + 1.
8.17 Evaluate
The Derivative
8.1 Using the A-definition, find the derivative /'(x) of the function /(x) = 2x - 7.
So,
Hence Thus, Answer
8.2 Using the A-definition, show that the derivative of any linear function /(x) = Ax + B is f ' ( x ) = A.
Then,
Hence. Thus,
8.3 Using the A-definition, find the derivative f ' ( x ) of the function /(x) = 2x2 - 3x + 5.
Thus, Hence,
lim
8.4 Using the A-definition, find the derivative /'(*) of the function f(x) = x3.
So,
So,
Thus,
8.5 State the formula for the derivative of an arbitrary polynomial function f(x) = anx" + an_lx" ' + • • • + a2x2 +
a1x+a0.
8.6 Write the derivative of the function f(x) = lx~ - 3x4 + 6x2 + 3x + 4.
/'(*) = 35x4 - I2x3 + \2x + 3.
8.7 Given functions f(x) and g(x), state the formulas for the derivatives of the sum f(x) + g(x), the product
fix) • e(x), and the quotient f(x) /g(x).
[Notice the various ways of denoting a derivative:
8.8 Using the product rule, find the derivative of f(x) = (Sx3 - 20* + 13)(4;t6 + 2x5 - lx2 + 2x).
F'(x) = (5x3-2Qx+13)(24x5 + Wx4-Ux + 2) + (4x" + 2x5-Ix2 + 2x)(15x2-20). [In such cases, do
not bother to carry out the tedious multiplications, unless a particular problem requires it.]
8.9 Using the formula from Problem 8-7. find the derivative of
49
, 50 CHAPTER 8
8.10 Using the formula from Problem 8.7, find the derivative of
8.11 Using the A-definition, find the derivative of
Hence
So,
and
8.12 Using formulas, find the derivatives of the following functions: (a)
(a) -40x4 + 3V3 x2 + 4Trx. Answer
(b) W2x50 + 36xu - 2Sx + i/7. Answer
8.13 Find the slope-intercept equation of the tangent line to the graph of the function f(x) = 4x3 - 7x2 at the point
corresponding to x = 3.
When x = 3, f(x) - 45. So, the point is (3,45). Recall that the slope of the tangent line is the derivative
/'(*), evaluated for the given value of x. But, /'(*) = 12x2 - Ux. Hence, /'(3) = 12(9) - 14(3) = 66.
Thus, the slope-intercept equation of the tangent line has the form y = 66x + b. Since the point (3,45) is on
the tangent line, 45 = 66(3) + 6, and, therefore, b = -153. Thus, the equation is v=66*-153.
Answer
8.14 At what point(s) of the graph of y = x5 + 4x - 3 does the tangent line to the graph also pass through the point
5(0,1)?
The derivative is y' = 5x4 + 4. Hence, the slope of the tangent line at a point A(xa, y0) of the graph is
5*o + 4. The line AB has slope So, the line AB is the tangent line if and
+4
only if (x0 + 4x0 - 4) Ix0 = 5x1 - Solving, x0 = — 1. So, there is only one point (—1, —8).
8.15 Specify all lines through the point (1, 5) and tangent to the curve y = 3>x3 + x + 4.
y' = 9x2 + l. Hence, the slope of the tangent line at a point (xa, ya) of the curve is 9*0 + 1. The slope of
the line through (x0, y0) and (1,5) is So, the tangent line passes
through (1,5) if and only if = 9x20 + l, 3*2 + j r 0 - l = (je 0 -l)(9*S + l), 3x30 + x0-l=9x30-
9**+ *„-!, 9*0 = 6*0, 6*o-9*o = 0, 3*0(2*0 - 3) = 0. Hence, *0 = 0 or * 0 = | , and the points on
the curve are (0, 4) and (§, ^). The slopes at these points are, respectively, 1 and f . So, the tangent lines are
y — 4 = x and y — *TT = T(X—%), or, equivalently, y = x + 4 and y = S f X — ".
8.16 Find the slope-intercept equation of the normal line to the graph of y = jc3 — x2 at the point where x = l.
The normal line is the line perpendicular to the tangent line. Since y' = 3x2 — 2x, the slope of the tangent
lineal x = 1 is 3(1)2 - 2(1) = 1. Hence, the slope of the normal line is the negative reciprocal of 1, namely
— 1. Thus, the required slope-intercept equation has the form y = —x + b. On the curve, when x = \,
y = (I) 3 - (I) 2 = 0. So, the point (1,0) is on the normal line, and, therefore, 0 = -1 + b. Thus, b = \,
and the required equation is y = — x + 1.
8.17 Evaluate
