Problem 46 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ g(\alpha)=5^{-\alpha / 2} \sin 2 \alpha $$

Step-by-Step Solution

Answer

The derivative of the given function is \(g'(\alpha)= -\frac{5^{-\alpha / 2} ln(5)\sin2\alpha}{2} + 5^{-\alpha / 2}\sin2\alpha \cos2\alpha\)

Step 1: Apply the Product Rule

Let's denote first function as \(u(\alpha)=5^{-\alpha / 2}\) and second function as \(v(\alpha)= \sin 2 \alpha\). Now, apply the Product Rule which gives \(g'(\alpha)= u'(\alpha)v(\alpha) + u(\alpha)v'(\alpha)\).

Step 2: Differentiate the functions

Now, differentiate the two functions (u and v separately). The derivative of the first function \(u'(\alpha)=\frac{d}{d\alpha}5^{-\alpha / 2}= -\frac{5^{-\alpha / 2} ln(5)}{2}\).As for the second function, \(v'(\alpha) = \frac{d}{d\alpha}\sin 2 \alpha = 2\cos2\alpha\).Substitute these into the product rule equation we got in step 1.

Step 3: Substitution and Simplification

Now, substitute these values into the equation from Step 1,\(g'(\alpha)= -\frac{5^{-\alpha / 2} ln(5)\sin2\alpha}{2} + \frac{2\cdot 5^{-\alpha / 2}\sin2\alpha \cos2\alpha}{2}\)

Step 4: Simplify Result

This simplifies to \(g'(\alpha)= -\frac{5^{-\alpha / 2} ln(5)\sin2\alpha}{2} + 5^{-\alpha / 2}\sin2\alpha \cos2\alpha\)

Key Concepts

Product Rule Differentiation

When you come across a function that is the product of two differentiable functions, the product rule is your go-to method for finding its derivative. The product rule states that if you have a function \( h(x) = f(x)g(x) \) where both \( f \) and \( g \) are differentiable, then the derivative of \( h \) with regard to \( x \) is \( h'(x) = f'(x)g(x) + f(x)g'(x) \).

In our exercise, we applied the product rule to the function \( g(\alpha) = 5^{-\alpha / 2} \sin 2\alpha \). We identified \( u(\alpha) = 5^{-\alpha / 2} \) and \( v(\alpha) = \sin 2\alpha \) as the two parts of the product. Using the product rule, we obtained the derivative of \( g \) by differentiating \( u \) and \( v \) separately and then combining them in the form \( u'(\alpha)v(\alpha) + u(\alpha)v'(\alpha) \) to get the result.

Chain Rule Calculus

The chain rule is a powerful technique in calculus for finding the derivative of a composite function. If you have a function \( y = g(f(x)) \), where \( g \) and \( f \) are both differentiable, then the chain rule tells you that the derivative of \( y \) with respect to \( x \) is \( dy/dx = g'(f(x)) \cdot f'(x) \).

In our example, we used the chain rule implicitly when differentiating \( u(\alpha) \), because this function is a composite of the exponential function and the scalar multiplication of \( \alpha \) by \( -1/2 \). We found the external derivative of the exponential function, then multiplied it by the derivative of the inner function \( -\alpha / 2 \) to properly apply the chain rule.

Trigonometric Functions Derivatives

Understanding how to differentiate trigonometric functions is essential in calculus. The main derivatives that are important to remember are:

• For \( \sin(x) \), the derivative is \( \cos(x) \).
• For \( \cos(x) \), the derivative is \( -\sin(x) \).
• For \( \tan(x) \), the derivative is \( \sec^2(x) \).

In our exercise, we differentiated \( v(\alpha) = \sin 2\alpha \), which requires an application of the chain rule as well. The derivative of \( \sin(x) \) is \( \cos(x) \) and considering the chain rule for the inner function \( 2\alpha \) gives us the result \( v'(\alpha) = 2\cos 2\alpha \) as seen in the solution.

Logarithmic Differentiation

Logarithmic differentiation is a method that can simplify finding the derivative of functions that are products, quotients, or powers where the variable appears as both a base and an exponent. The technique involves taking the natural logarithm \( \ln \) of both sides of the equation, applying the properties of logarithms to simplify, differentiating implicitly with respect to the variable, and then solving for the derivative.

In our step-by-step solution, we dealt with the function \( u(\alpha) = 5^{-\alpha / 2} \). When finding its derivative, \( u'(\alpha) \), we used logarithmic differentiation in a subtle way. By recognizing that the natural logarithm of the base, \( \ln(5) \), can be used to simplify the exponentiation process when differentiating. This approach is a valuable tool for managing more complex derivatives involving exponentials and logarithms.