Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 3 - Derivatives

Problema 3.33

Capitolo 3, Problema 3.33

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

Guida verificata passo dopo passo

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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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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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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Simplifying Trig Expressions

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