Problem 39 - Calculus Early Transcendentals

Question

Derivatives Evaluate the derivatives of the following functions. \(G(y)=y^{\sin y}\)

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(G(y) = y^{\sin y}\). Answer: The derivative of the function \(G(y) = y^{\sin y}\) is \((G(y))' = y^{\sin y-1} (\sin y + y \cos y)\).

Step 1: Identify the base and exponent in the function

For our function \(G(y)=y^{\sin y}\), we have the following: - Base: \(y\) - Exponent: \(\sin y\)

Step 2: Apply the chain rule for exponents

To find the derivative of an exponent function, we need to apply the chain rule: \((u^v)'= vu^{v-1} \cdot u'\) where \(u=y\) and \(v=\sin y\).

Step 3: Calculate the derivatives of base and exponent

Now, we need to find the derivatives of the base, \(u'\), and exponent, \(v'\). For the base, \(u=y\), its derivative is simply: \(u' = \frac{dy}{dy} = 1\) For the exponent, \(v=\sin y\), its derivative is: \(v' = \frac{d(\sin y)}{dy} = \cos y\)

Step 4: Apply the chain rule and replace \(u'\) and \(v'\)

Applying the chain rule and replacing \(u'\) and \(v'\), we get: \((G(y))' = (\sin y)\cdot y^{\sin y - 1} \cdot 1 + y^{\sin y} \cdot \cos y\)

Step 5: Simplify the expression

Now simplify the derivative expression: \((G(y))' = y^{\sin y-1} (\sin y + y \cos y)\) So the derivative of \(G(y) = y^{\sin y}\) is: \((G(y))' = y^{\sin y-1} (\sin y + y \cos y)\)

Key Concepts

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The exponent indicates how many times the base is multiplied by itself. In the function given, \( G(y) = y^{\sin y} \), \(y\) serves as the base, and \(\sin y\) is the exponent.
Understanding exponentiation in differentiation is crucial because even when both base and exponent include variables, you must approach it with special rules, such as the chain rule which we'll explain later. Generally, the derivative of a function \(x^k\), where \(k\) is constant, is given by the formula \(k \cdot x^{k-1}\). However, when the exponent itself is a variable, the process involves more steps. This is where knowledge of exponent functions broadens: instead of separating into neat constant parts, you incorporate varying components into your derivative calculations.

When working with exponents in derivatives:

• Identify both base and exponent separately.
• Apply appropriate mathematical rules depending on whether the exponent is fixed or variable.
• Be aware of trigonometric functions as exponents can change how derivatives are approached.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It is essential when dealing with functions where an exponent is not just a simple number, but rather a function of the variable, like in this exercise with \( \sin y \) as the exponent.
The chain rule states that if you have two functions composed together, say \( f(g(x)) \), the derivative is given by \( f'(g(x)) \cdot g'(x) \). This rule allows us to "chain" the derivatives of the inside and outside functions. In the context of our problem where the function is in the form \( y^{\sin y} \), it involves applying the chain rule:

• Differentiate the outer function considering the inner function as a constant.
• Multiply that result by the derivative of the inner function itself.

In our example, when applying the chain rule for \( u^v \), we get:

• The outside function is \(u^v\), resulting in \(vu^{v-1}\).
• The derivative from the inside function involves the derivative of \(\sin y\), which is \(\cos y\).

Using the chain rule efficiently is about understanding how to maneuver through both the outer function's structure and the subtleties of its inner workings, such as the variable exponent here.

Calculus

Calculus is the branch of mathematics that studies change, a route created by Newton and Leibniz. It consists of two principal areas: differentiation and integration. Differentiation focuses on rates of change, resembling the slope of curves at specific points. With derivatives, you find how the output of a function varies as its input changes.
The exercise example \(G(y) = y^{\sin y}\) revolved around differentiation using calculus principles like the chain rule and exponentiation derivative rules to find \((G(y))'\).
In calculus:

• Differentiation helps in understanding and calculating rates of change and slopes of curves.
• Advance through complex equations by breaking them into smaller, manageable derivative pieces.
• Use derivatives to analyze function behaviors, like maxima, minima, and inflection points.

Differentiation is an essential skill within calculus and mastering it widens your ability to tackle various mathematical challenges, whether in pure mathematics or in applied sciences. By practicing these techniques, you not only solve problems in the classroom but also develop sharper analytical skills.