Question

Find the indicated derivative. $$ \frac{d y}{d t}, \text { where } y=\left[\sin t \tan \left(t^{2}+1\right)\right] $$

Short answer

\( \frac{d y}{d t} = \cos t \tan(t^2 + 1) + 2t \sin t \sec^2(t^2 + 1) \)

Step-by-step solution

Apply the Product Rule

The function given is a product of two functions: \( \sin t \) and \( \tan(t^2 + 1) \). According to the product rule, if \( y = u imes v \), then \[ \frac{d y}{d t} = u'v + uv' \]Identify \( u = \sin t \) and \( v = \tan(t^2 + 1) \).

Differentiate \( u = \sin t \)

Find the derivative of \( u \) with respect to \( t \). \[ \frac{d u}{d t} = \cos t \]

Differentiate \( v = \tan(t^2 + 1) \) using the Chain Rule

First, let \( g(t) = t^2 + 1 \). Thus, \( v = \tan(g(t)) \).The chain rule states that \( \frac{d v}{d t} = \frac{d}{d t} [\tan(g(t))] = \sec^2(g(t)) \cdot \frac{d g}{d t} \)Calculate \( \frac{d g}{d t} \): \\[ \frac{d g}{d t} = 2t \] Substitute back into the chain rule expression:\[ \frac{d v}{d t} = \sec^2(t^2 + 1) imes 2t \]

Substitute into the Product Rule Formula

Now substitute \( \frac{d u}{d t} = \cos t \), \( u = \sin t \), \( \frac{d v}{d t} = 2t \sec^2(t^2 + 1) \), and \( v = \tan(t^2 + 1) \) into the product rule formula:\[ \frac{d y}{d t} = \cos t \cdot \tan(t^2 + 1) + \sin t \cdot (2t \sec^2(t^2 + 1)) \]Simplify the expression:\[ \frac{d y}{d t} = \cos t \tan(t^2 + 1) + 2t \sin t \sec^2(t^2 + 1) \]

Final Derivative

The final expression, which is the derivative of the function \( y = \sin t \tan(t^2 + 1) \) with respect to \( t \), is:\[ \frac{d y}{d t} = \cos t \tan(t^2 + 1) + 2t \sin t \sec^2(t^2 + 1) \]

Key concepts

Derivative

The derivative of a function measures how the function's output changes concerning its input. Think of it as the 'slope' of the function at any given point. In basic terms, if you were on a curve, the derivative tells you how steep you are. When working with derivatives, we indicate the variable of differentiation. For instance, if we are finding the derivative of a function \( y \) with respect to \( t \), we denote it as \( \frac{d y}{d t} \). This reveals how \( y \) changes when \( t \) changes slightly.
In calculus, derivatives are essential in understanding how variables interact differentially. They help in creating models that predict changes and optimization in real-world scenarios. Key to this is the derivative's ability to expose minute alterations within functions, providing insights into their behavior.

Chain Rule

The Chain Rule is a technique in calculus used to differentiate composite functions. A composite function is where one function is nested inside another. For instance, consider \( v = \tan(t^2 + 1) \). Here, \( t^2 + 1 \) is an inner function, and \( \tan \) is an outer function. Using the Chain Rule, we can break this down to find the derivative.
To apply the Chain Rule, first identify the outer function and differentiate it with respect to its inner function. Next, differentiate the inner function with respect to the original variable. Multiply these derivatives together to get the result.

• Differentiate the outer function: \( \tan(g(t)) \) becomes \( \sec^2(g(t)) \)
• Differentiate the inner function: \( g(t) = t^2 + 1 \) becomes \( 2t \)
• Combine with the Chain Rule: \( \frac{d}{d t} [\tan(t^2 + 1)] = \sec^2(t^2 + 1) \times 2t \)

The Chain Rule is essential for making complex differentiation problems more manageable by offering a systematic approach.

Differentiation

Differentiation is the core operation in calculus that captures how a quantity changes. It transforms a function into its derivative, granting insight into its rate of change. Here, the function \( y = \sin t \tan(t^2 + 1) \) needs differentiation to find how \( y \) varies as \( t \) changes.
The Product Rule is a specific differentiation technique applied when dealing with products of functions. If a function is the product of two smaller functions, like \( u = \sin t \) and \( v = \tan(t^2 + 1) \), then the Product Rule tells us:
\[ \frac{d y}{d t} = u'v + uv' \]

• Compute the derivative of \( u \): \( \frac{d u}{d t} = \cos t \)
• Use the Chain Rule for \( v \): \( \frac{d v}{d t} = 2t \sec^2(t^2 + 1) \)
• Substitute back: \( \frac{d y}{d t} = \cos t \tan(t^2 + 1) + 2t \sin t \sec^2(t^2 + 1) \)

This rule provides a structured means to differentiate products, simplifying complex calculations.

Trigonometric Functions

Trigonometric functions like \( \sin \), \( \cos \), and \( \tan \) are periodic functions central to many areas of mathematics, particularly calculus. They help model oscillations, waves, and cycles found in natural phenomena and engineered systems.
Understanding how to differentiate these functions is vital. The basic derivatives are:

• \( \frac{d}{d t}(\sin t) = \cos t \)
• \( \frac{d}{d t}(\cos t) = -\sin t \)
• \( \frac{d}{d t}(\tan t) = \sec^2 t \)

In the given problem \( y = \sin t \tan(t^2 + 1) \), we see \( \sin t \) and \( \tan(t^2 + 1) \). We apply the differentiation techniques to find how these functions blend in calculus, accounting for both their individual behavior and their interaction through products or chaining. Mastery of such differentiation transforms tasks involving trigonometric functions from daunting to doable.