3.3.1 The Derivative of the Reciprocal Function

Question 1

derivative of the reciprocal function

Answer 1

• The derivative of is . f(x) = x^-1 is f’(x) = -x^-2
• To find the equation of a line tangent to a curve, take the derivative, evaluate the derivative at the point of tangency to find the slope, and substitute the slope and the point of tangency into the point-slope form of a line.

Question 2

note

Answer 2

• The function f(x)=x^-1 is called the reciprocal function because for any value of x the function produces the reciprocal of x as its output.
• Remember, 1/x^n can be expressed as x^-n. To find the equation of a line tangent to the reciprocal function at a point, start by finding the derivative. Notice that you must find a common denominator for the numerator in order to evaluate the limit.
• By combining the terms in the numerator, it is possible to simplify the expression and cancel some x-terms.
• The derivative of is f(x) = x^-1 is f’(x) = -x^-2.
• Once you have found the derivative of the reciprocal function, evaluate the derivative at the point of tangency to find the slope of the tangent line.
• Use the slope and the point of tangency to express the equation of the line in point-slope form.

Question 3

Suppose f(x)=4/x. What is the slope of the line tangent to f when x=3?

Answer 3

−4/9

Question 4

Find the derivative of f(x)=−1/√5 x.

Answer 4

f′(x)=1/√5 x^2

Question 5

At what points does the equation of the line tangent to the curve y=1/x have a slope equal to −1?

Answer 5

(−1, −1) and (1, 1)

Question 6

Find the equation of the line tangent to the curve y = 1 / (2x) when x = 1.

Answer 6

y = (−1/2) x + 1

Question 7

Suppose f (x) = −4 / x. Find the equation of the line tangent to f (x) at (4, −1).

Answer 7

y = x / 4 − 2

Question 8

Find the derivative of f(x)=−1/√2 x +2x.

Answer 8

f′(x)=2+1/√2 x^2

Question 9

Suppose f (x) = 6 / x. What is the slope of the line tangent to f when x = −2?

Answer 9

−3/2

Question 10

Find the derivative of f(x)=1/4x.

Answer 10

f′(x)=−1/4x^2

Question 11

Suppose a particle’s position is given by f (t) = −4 / t, where t is measured in seconds and f (t) is given in centimeters. What is the velocity of the particle when t = 1?

Answer 11

4 cm / sec

Question 12

Find the equation of the line tangent to the curve y = 6 / x when x = 3.

Answer 12

y−2=−2/3x+2

Question 13

Find the derivative of f(x)=2/3x.

Answer 13

f′(x)=−2/3x^2

Question 14

Suppose f (x) = −2 / x. Find the equation of the line tangent to f (x) at (1, −2).

Answer 14

y = 2x − 4

Question 15

Suppose a particle’s position is given by f (t) = −2 / t, where t is measured in seconds and f (t) is given in centimeters. At what time is the velocity of the particle equal to 4 cm / s?

Answer 15

t=1/√2

Question 16

Find the derivative of f(x)=1/3x.

Answer 16

f′(x)=−1/3x^2

Question 17

At what points does the equation of the line tangent to the curve y=1/x+3 have a slope equal to −1?

Answer 17

(−1, 2) and (1, 4)

Question 18

Suppose a particle’s position is given by f (t) = 2 / t, where t is measured in seconds and f (t) is given in centimeters. What is the velocity of the particle when t = 2?

Answer 18

−1/2 cm / s

Question 19

Suppose a particle’s position is given by f(t)=3/t,where t is measured in seconds and f(t) is given in centimeters. At what time is the velocity of the particle equal to −12?

Answer 19

t=1/2