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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 3.33
Chapter 3, Problem 3.33

Find the derivatives of the functions in Exercises 1–42.
_____
𝔂 = / x² + x
√ x²

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1
Step 1: Simplify the given function. The function is 𝔂 = x² + x√x². Notice that x√x² can be rewritten as x * x, which simplifies to x². Therefore, the function becomes 𝔂 = x² + x².
Step 2: Combine like terms. Since both terms are x², the function simplifies to 𝔂 = 2x².
Step 3: Differentiate the simplified function. To find the derivative of 𝔂 = 2x², apply the power rule. The power rule states that the derivative of x^n is n*x^(n-1).
Step 4: Apply the power rule to each term. For the term 2x², the derivative is 2 * 2 * x^(2-1), which simplifies to 4x.
Step 5: Write the final expression for the derivative. The derivative of the function 𝔂 = 2x² is 𝔂' = 4x.

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Key Concepts

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

Derivatives

The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the curve of the function at any given point. Derivatives can be computed using various rules, such as the power rule, product rule, and quotient rule, depending on the form of the function.
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Derivatives

Quotient Rule

The quotient rule is a method for finding the derivative of a function that is the ratio of two other functions. If you have a function defined as f(x) = g(x)/h(x), the derivative f'(x) is given by (g'(x)h(x) - g(x)h'(x)) / (h(x))². This rule is essential when differentiating functions that involve division, as seen in the given problem.
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The Quotient Rule

Simplifying Functions

Before differentiating complex functions, it is often helpful to simplify them. This can involve factoring, combining like terms, or rewriting expressions in a more manageable form. In the context of the given function, simplifying the expression can make it easier to apply the derivative rules accurately and efficiently.
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Related Practice
Textbook Question

Find the first and second derivatives of the functions in Exercises 33–38.


w = ((1 + 3z) / 3z) (3 − z)

Textbook Question

A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?

Textbook Question

Derivative Calculations


In Exercises 1–12, find the first and second derivatives.


y = 6x² − 10x − 5x⁻²

Textbook Question

Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%?

Textbook Question

One-Sided Derivatives


Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.

Textbook Question

Second Derivatives


In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.


y² – 2x = 1 – 2y