Problem 5 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\csc x-4 \sqrt{x}+\frac{7}{e^{x}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = -\csc x \cot x - 2x^{-1/2} - 7e^{-x} \).

Step 1: Differentiate the Cosecant Function

The first term in the expression is \( y = \csc x \). Recall that the derivative of \( \csc x \) with respect to \( x \) is \( -\csc x \cot x \). So, the derivative of \( \csc x \) is \( -\csc x \cot x \).

Step 2: Differentiate the Square Root Function

Next, take the derivative of \( -4\sqrt{x} \). Rewrite \( \sqrt{x} \) as \( x^{1/2} \). Using the power rule \( \frac{d}{dx}x^n = nx^{n-1} \), differentiate to get \( \frac{d}{dx}(-4x^{1/2}) = -4 \times \frac{1}{2}x^{-1/2} = -2x^{-1/2} \).

Step 3: Differentiate the Exponential Function

For the last term, we have \( \frac{7}{e^x} \). This can be rewritten as \( 7e^{-x} \). The derivative of \( e^{-x} \) is \( -e^{-x} \). Using this, differentiate to get \( \frac{d}{dx}(7e^{-x}) = 7\cdot(-e^{-x}) = -7e^{-x} \).

Step 4: Combine All Derivatives

Combine the derivatives for all the terms to get the final expression for \( \frac{dy}{dx} \):\[\frac{dy}{dx} = -\csc x \cot x - 2x^{-1/2} - 7e^{-x}\]

Key Concepts

Derivative of Trigonometric Functions

The derivative of trigonometric functions is a key concept in calculus, important for analyzing how these functions change. In our exercise, we start by differentiating the function \( y = \csc x \), which is the cosecant function. The cosecant function is the reciprocal of the sine function. Therefore, its derivative is unique. The derivative of \( \csc x \) is \( -\csc x \cot x \). This result stems from the need to apply both the quotient rule and knowledge of how the sine and cosine functions behave.

Here are a few important points to remember:

• \( \csc x = \frac{1}{\sin x} \)
• \( \cot x = \frac{\cos x}{\sin x} \)
• Derivative of \( \sin x \) is \( \cos x \) and derivative of \( \cos x \) is \( -\sin x \)

Using these identities can help understand why the derivative of \( \csc x \) includes both the cosecant and cotangent functions.

Power Rule in Calculus

The power rule is one of the simplest and most essential tools for finding derivatives in calculus. It allows us to differentiate functions with any power of \( x \) quickly. In our case, we need to differentiate \( -4\sqrt{x} \), which can be rewritten using exponents as \( -4x^{1/2} \).

Applying the power rule, \( \frac{d}{dx} x^n = nx^{n-1} \), gives us:

• The exponent \( 1/2 \) moves to the front
• Subtract one from the exponent \( 1/2 - 1 = -1/2 \)
• Multiply the constant (\-4) by the new exponent

This leads to the derivative: \( -4 \times \frac{1}{2}x^{-1/2} = -2x^{-1/2} \). Using the power rule helps simplify and make sense of derivatives quickly, especially with more complicated expressions.

Derivative of Exponential Functions

Exponential functions have unique properties when it comes to differentiation. The function \( e^x \) is special because its derivative is itself: \( \frac{d}{dx} e^x = e^x \). In our exercise, however, we have \( \frac{7}{e^x} \), which can be rewritten as \( 7e^{-x} \). This requires applying the chain rule in calculus.

For an exponential function with a negative exponent \( e^{-x} \), the derivative is \(-e^{-x}\). Hence, differentiating \( 7e^{-x} \) involves:

• Multiplying the constant 7 by the derivative of \( e^{-x} \)
• The result is \( -7e^{-x} \)

Understanding this process is key when dealing with exponential functions in calculus. It highlights how constants affect the differentiation and the importance of recognizing underlying exponent rules.