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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.5.39

23–51. Calculating derivatives Find the derivative of the following functions.
y = sin x / 1 + cos x

Guida verificata passo dopo passo
1
Step 1: Recognize that the function \( y = \frac{\sin x}{1 + \cos x} \) is a quotient of two functions, \( u(x) = \sin x \) and \( v(x) = 1 + \cos x \). To find the derivative, we will use the quotient rule.
Step 2: Recall the quotient rule for derivatives, which states that if \( y = \frac{u(x)}{v(x)} \), then \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).
Step 3: Differentiate the numerator \( u(x) = \sin x \). The derivative \( u'(x) = \cos x \).
Step 4: Differentiate the denominator \( v(x) = 1 + \cos x \). The derivative \( v'(x) = -\sin x \).
Step 5: Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula to find \( y' \).

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Derivatives

A derivative represents the rate of change of a function with respect to its variable. It is a fundamental concept in calculus that allows us to determine how a function behaves at any given point. The derivative can be interpreted as the slope of the tangent line to the curve of the function at a specific point.
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Quotient Rule

The Quotient Rule is a formula used to find the derivative of a function that is the ratio of two other functions. If you have a function y = u/v, where u and v are both differentiable functions, the derivative is given by (v * du/dx - u * dv/dx) / v^2. This rule is essential for differentiating functions like the one in the question.
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Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in calculus and describe relationships between angles and sides of triangles. Their derivatives are also crucial, as they follow specific rules: the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x). Understanding these derivatives is key to solving problems involving trigonometric functions.
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Introduction to Trigonometric Functions