Question

Compute the derivative of the given function. $$g(t)=\cos \left(t^{2}+3 t\right) \sin (5 t-7)$$

Short answer

The derivative is \(-\sin(t^2 + 3t)(2t + 3)\sin(5t - 7) + 5\cos(t^2 + 3t)\cos(5t - 7)\)."

Step-by-step solution

Identify the Function Type

The function given is a product of two functions, specifically, it is in the form \( u(t)v(t) \) where \( u(t) = \cos(t^2 + 3t) \) and \( v(t) = \sin(5t - 7) \). This means we'll use the product rule for differentiation.

Differentiate using the Product Rule

The product rule states that if \( h(t) = u(t) \, v(t) \), then \( h'(t) = u'(t) \, v(t) + u(t) \, v'(t) \). We'll need to find the derivatives \( u'(t) \) and \( v'(t) \).

Differentiate \( u(t) = \cos(t^2 + 3t) \)

The derivative of \( \cos(x) \) is \(-\sin(x) \). Using the chain rule, the derivative of \( \cos(t^2 + 3t) \) is \(-\sin(t^2 + 3t) \cdot (2t + 3)\).

Differentiate \( v(t) = \sin(5t - 7) \)

The derivative of \( \sin(x) \) is \( \cos(x) \). Therefore, using the chain rule, the derivative of \( \sin(5t - 7) \) is \( \cos(5t - 7) \cdot 5 \).

Apply the Product Rule Formula

Now, substituting these derivatives into the product rule formula, we get: \[ g'(t) = [-\sin(t^2 + 3t) \cdot (2t + 3)] \cdot \sin(5t - 7) + \cos(t^2 + 3t) \cdot [\cos(5t - 7) \cdot 5]. \]

Key concepts

Derivative

In calculus, the derivative represents the rate at which a function changes with respect to a variable. It is like a snapshot of change at a particular point. When dealing with a function, \(g(t)\), the derivative, denoted by \(g'(t)\), tells us how quickly the output values change as we make small changes to the input variable \(t\). This concept is critical in understanding the behavior of functions.
To find a derivative, we use differentiation techniques. For basic functions like a constant or \(x^n\), standard rules apply. However, when functions combine components, like through addition, multiplication, or composition, we must use specific rules, such as the product or chain rule, to differentiate effectively.
For complex functions, conceptualizing derivatives in a structured way allows us to predict trends, maximize or minimize functions, and understand real-world phenomena. Derivatives require careful step-by-step application of these rules, ensuring each part is correctly processed to arrive at the correct result.

Product Rule

The product rule is vital when differentiating functions that are products of two or more functions. Suppose \(h(t) = u(t) \, v(t)\); the product rule provides a formula to find its derivative: \(h'(t) = u'(t) \, v(t) + u(t) \, v'(t)\). This method ensures both components contribute accurately to the derivative.
Here's how to apply the product rule:

• Identify the individual functions being multiplied, consider what \(u(t)\) and \(v(t)\) are.
• Differentiate each part separately, finding \(u'(t)\) and \(v'(t)\).
• Apply the product rule formula to assemble the derivative.

In our original problem, \(u(t) = \cos (t^2 + 3t)\) and \(v(t) = \sin (5t - 7)\). This example shows how breaking down components allows us to apply the derivative in manageable parts, making it less daunting and clearer.

Chain Rule

The chain rule is used for finding the derivative of a composite function, which involves a function within another function. It's essential when a function involves nested operations, like \(\cos(t^2 + 3t)\). The rule is given by: if \(y = f(g(x))\), then the derivative is \(y' = f'(g(x)) \cdot g'(x)\).
For example, in differentiating \(\cos(t^2 + 3t)\), \(t^2 + 3t\) is a function inside \(\cos\). The outside function, \(\cos(x)\), differentiates to \(-\sin(x)\). The inside function differentiates to \(2t + 3\). By the chain rule, these results multiply to \(-\sin(t^2 + 3t) \cdot (2t + 3)\).
Here's how to think through the chain rule:

• Recognize the outer function and differentiate it, keeping the inner function unchanged.
• Differentiate the inner function itself.
• Multiply these derivatives to find the composite function's derivative.

It's a powerful technique for managing complex differentiations easily.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in calculus for modeling periodic behavior. When differentiating these functions, specific rules govern the process: \(\sin(x)\) differentiates to \(\cos(x)\), and \(\cos(x)\) differentiates to \(-\sin(x)\). Understanding this relationship helps in computing derivatives of more complex trigonometric expressions.
In the given problem, we worked with both \(\sin\) and \(\cos\): \(\cos(t^2 + 3t)\) and \(\sin(5t - 7)\). Each trigonometric function's derivative includes not just the simple change but often involves using the chain rule to consider the inside variable changes.
Tips for working with trigonometric functions:

• Always remember their basic derivatives: sine to cosine, cosine to negative sine.
• Apply the chain rule if these functions include complex expressions inside.
• Carefully track negative signs, as they are easy to lose or misplace.

With practice, handling these functions becomes intuitive, enabling you to solve even more intricate calculus problems.