In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
19. y = t arctan(t) - 1/2 ln(t)

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Identify the function to differentiate: \(y = t \arctan(t) - \frac{1}{2} \ln(t)\), where the variable is \(t\).

Apply the derivative operator \(\frac{dy}{dt}\) to each term separately: \(\frac{d}{dt} \left( t \arctan(t) \right)\) and \(\frac{d}{dt} \left( -\frac{1}{2} \ln(t) \right)\).

For the first term, use the product rule since it is a product of two functions of \(t\): \(u = t\) and \(v = \arctan(t)\). Recall the product rule: \(\frac{d}{dt}(uv) = u'v + uv'\).

Calculate the derivatives of the individual parts: \(u' = \frac{d}{dt}(t) = 1\) and \(v' = \frac{d}{dt}(\arctan(t)) = \frac{1}{1 + t^2}\).

For the second term, use the derivative of the natural logarithm: \(\frac{d}{dt} \left( -\frac{1}{2} \ln(t) \right) = -\frac{1}{2} \cdot \frac{1}{t}\). Combine all parts to write the full derivative expression.

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주요 개념

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Derivative of Inverse Trigonometric Functions

The derivative of the arctan function, arctan(t), is 1/(1 + t²). Understanding how to differentiate inverse trigonometric functions is essential for handling terms like t arctan(t) in the expression.

Product Rule

The product rule is used to differentiate products of two functions. For y = u(t)v(t), the derivative is u'v + uv'. This rule applies to the term t arctan(t), where both t and arctan(t) are functions of t.

Derivative of Logarithmic Functions

The derivative of the natural logarithm ln(t) is 1/t. Recognizing this allows for straightforward differentiation of the term -1/2 ln(t), which is part of the given function.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.19. y = t arctan(t) - 1/2 ln(t)