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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 3 - DerivativesProblema 3.7.29
Capitolo 3, Problema 3.7.29

27–76. Calculate the derivative of the following functions.
y=10x+1y=\(\sqrt{10x+1}\)

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1
Step 1: Recognize that the function given is \( y = \sqrt{10x + 1} \). This can be rewritten using exponent notation as \( y = (10x + 1)^{1/2} \).
Step 2: To find the derivative, apply the chain rule. The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative \( y' = f'(g(x)) \cdot g'(x) \).
Step 3: Identify the outer function \( f(u) = u^{1/2} \) and the inner function \( g(x) = 10x + 1 \).
Step 4: Differentiate the outer function with respect to \( u \): \( f'(u) = \frac{1}{2}u^{-1/2} \).
Step 5: Differentiate the inner function with respect to \( x \): \( g'(x) = 10 \).

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Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In practical terms, the derivative provides the slope of the tangent line to the curve of the function at any given point.
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Derivatives

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It states that if a function y is composed of two functions u and x (i.e., y = f(u) and u = g(x)), then the derivative of y with respect to x can be found by multiplying the derivative of f with respect to u by the derivative of g with respect to x. This rule is essential for handling functions that are nested within each other.
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Intro to the Chain Rule

Square Root Function

The square root function, denoted as √x, is a function that returns the non-negative value whose square is x. When differentiating a square root function, it is important to apply the power rule, as the square root can be expressed as x^(1/2). Understanding how to differentiate square root functions is crucial for solving problems involving derivatives of such expressions.
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Multiplying & Dividing Functions
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