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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.R.17

9–61. Evaluate and simplify y'.


y = 5t² sin t

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First, identify the given function y = 5t² sin(t). We need to find the derivative y' with respect to t.
Apply the product rule for differentiation, which states that if you have a function u(t) * v(t), its derivative is u'(t) * v(t) + u(t) * v'(t). Here, u(t) = 5t² and v(t) = sin(t).
Differentiate u(t) = 5t² with respect to t. The derivative is u'(t) = 10t.
Differentiate v(t) = sin(t) with respect to t. The derivative is v'(t) = cos(t).
Combine the results using the product rule: y' = u'(t) * v(t) + u(t) * v'(t) = 10t * sin(t) + 5t² * cos(t). Simplify the expression if possible.

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Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change of the function with respect to its variable. In this case, we need to differentiate the function y = 5t² sin t to find y'.
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Product Rule

The Product Rule is a specific rule used in differentiation when dealing with the product of two functions. It states that if you have two functions u(t) and v(t), the derivative of their product is given by u'v + uv'. In the given function y = 5t² sin t, we will apply the Product Rule to differentiate the product of 5t² and sin t.
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Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental functions in calculus that describe relationships in triangles and periodic phenomena. The derivative of sin t is cos t, which is essential for differentiating the sin t component in the function y = 5t² sin t. Understanding how to differentiate these functions is crucial for solving the problem.
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Introduction to Trigonometric Functions