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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3
Chapter 3, Problem 3

Finding Derivative Functions and Values 


Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.


g(t) = 1/t²; g′(−1), g′(2), g′(√3)

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1
Start by recalling the definition of the derivative: the derivative of a function g(t) at a point t is given by the limit g'(t) = lim(h→0) [(g(t+h) - g(t))/h].
Substitute the given function g(t) = 1/t² into the definition of the derivative. This gives g'(t) = lim(h→0) [(1/(t+h)² - 1/t²)/h].
Simplify the expression inside the limit. Find a common denominator for the fractions in the numerator: (1/(t+h)² - 1/t²) = (t² - (t+h)²) / (t²(t+h)²).
Expand and simplify the numerator: t² - (t+h)² = t² - (t² + 2th + h²) = -2th - h². Substitute this back into the limit expression.
Evaluate the limit: g'(t) = lim(h→0) [(-2th - h²) / (h * t²(t+h)²)]. Cancel h from the numerator and denominator, then take the limit as h approaches 0 to find g'(t). Finally, substitute t = -1, t = 2, and t = √3 to find g'(-1), g'(2), and g'(√3) respectively.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definition of Derivative

The derivative of a function at a point is the limit of the average rate of change of the function as the interval approaches zero. Mathematically, it is defined as f'(x) = lim(h→0) [f(x+h) - f(x)]/h. This concept is crucial for understanding how to calculate the instantaneous rate of change of a function.
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Power Rule for Derivatives

The power rule is a basic derivative rule used to find the derivative of functions in the form of f(x) = x^n. According to the power rule, the derivative is f'(x) = n*x^(n-1). This rule simplifies the process of finding derivatives for polynomial functions, including rational functions like g(t) = 1/t².
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Substitution in Derivatives

Once the derivative function is found, substitution involves plugging specific values into the derivative to find the rate of change at those points. For example, after finding g'(t), substitute t = -1, t = 2, and t = √3 to find g'(-1), g'(2), and g'(√3), respectively. This step is essential for evaluating the derivative at given points.
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Related Practice
Textbook Question

Find dy/dt when x = 1 if y = x² + 7x − 5 and dx/dt = ¹/₃.

Textbook Question

Theory and Examples


Intersecting normal line The line that is normal to the curve x² + 2xy – 3y² = 0 at (1,1) intersects the curve at what other point?

Textbook Question

In Exercises 41–58, find dy/dt.


y = √(3t + (√2 + √(1 − t)))

Textbook Question

Differentiating Implicitly


Use implicit differentiation to find dy/dx in Exercises 1–14.


x⁴ + sin y = x³y²

Textbook Question

Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.


x ƒ(x) g(x) ƒ'(x) g'(x)

0 1 1 -3 1/2

1 3 5 1/2 -4


Find the first derivatives of the following combinations at the given value of x.


a. 6ƒ(x) - g(x), x = 1

1
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Textbook Question

Find the derivatives of the functions in Exercises 19–40.


p = √(3 − t)