Question

If \(y=\frac{3}{x^{2}}-2 \sin 4 x+\frac{2}{\mathrm{e}^{x}}+\ln 5 x\), determine \(\frac{\mathrm{d} y}{\mathrm{~d} x}\)

Short answer

\(-\frac{6}{x^3} - 8\cos(4x) - 2e^{-x} + \frac{1}{x}\).

Step-by-step solution

Differentiate the first term

The first term to differentiate is \(\frac{3}{x^2}\). We can rewrite it as \(3x^{-2}\). Using the power rule, which states \(\frac{d}{dx}(x^n) = nx^{n-1}\), we differentiate: \(\frac{d}{dx}(3x^{-2}) = 3(-2)x^{-3} = -\frac{6}{x^3}\).

Differentiate the second term

The second term is \(-2\sin(4x)\). The derivative of \(\sin(4x)\) is \(4\cos(4x)\) by the chain rule. Thus, the differentiation gives: \(\frac{d}{dx}(-2\sin(4x)) = -2 \cdot 4\cos(4x) = -8\cos(4x)\).

Differentiate the third term

The third term is \(\frac{2}{e^x}\). This can be rewritten as \(2e^{-x}\). Using the power rule and the chain rule, differentiate: \(\frac{d}{dx}(2e^{-x}) = 2(-e^{-x}) = -2e^{-x}\).

Differentiate the fourth term

The fourth term is \(\ln(5x)\). The derivative of \(\ln(u)\) is \(\frac{u'}{u}\), so the derivative of \(\ln(5x)\) is \(\frac{5}{5x} = \frac{1}{x}\).

Combine the derivatives

Now, we combine the derivatives of each term from the previous steps to find \(\frac{dy}{dx}\). Summing them gives: \(-\frac{6}{x^3} - 8\cos(4x) - 2e^{-x} + \frac{1}{x}\).

Key concepts

Power Rule

The Power Rule is a fundamental tool in differentiation, especially when dealing with polynomial functions. It's simple, yet powerful, providing a straightforward way to find the derivative of expressions involving powers of a variable. The Power Rule states that if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\).

This means you multiply the exponent by the base and then decrease the exponent by one. Consider the exercise problem: the term \(\frac{3}{x^2}\) was rewritten as \(3x^{-2}\). Using the Power Rule, the derivative is calculated as \(3 \times (-2)x^{-3} = -\frac{6}{x^3}\).

Remember, applying the Power Rule effectively transforms the exponent operation into a simpler form, making it an essential technique for solving derivatives.

Chain Rule

The Chain Rule is essential for differentiating composite functions, where one function is nested inside another. It allows us to systematically "unpack" these functions and differentiate them more easily.

The Chain Rule states that if you have a composite function \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\). This means you first differentiate the outer function and multiply it by the derivative of the inner function.

In our exercise, for the term \(-2\sin(4x)\), we used the Chain Rule. We treat \(\sin(4x)\) as \(\sin(u)\) where \(u = 4x\). The derivative of \(\sin(u)\) is \(\cos(u)\) and \(du/dx = 4\). Hence, \(-2\sin(4x)\) becomes \(-2 \cdot 4\cos(4x) = -8\cos(4x)\).

• Break down the composite function into layers.
• Differentiating from the outer layer inwards.
• Multiply the derivatives as per the rule.

This method is crucial when handling more complex functions that involve innately linked parts.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique for differentiating functions where direct application of standard differentiation rules would be complex or unwieldy. This method is particularly useful for functions involving products, quotients, or powers of functions.

To perform logarithmic differentiation, you typically take the natural logarithm of both sides of the equation \(y = f(x)\). From there, you differentiate the resulting equation using properties of logarithms, and solve for \(\frac{dy}{dx}\).

In our exercise, the term \(\ln(5x)\) is an example suitable for logarithmic differentiation. By recognizing \(\ln(u)\)'s derivative is \(u'/u\), where \(u=5x\), we find \(\frac{d}{dx}(\ln(5x)) \) which simplifies to \(\frac{1}{x}\) after recognizing that the derivative of \(5x\) over \(5x\) simplifies to \(5/5x = 1/x\).

This approach effectively reduces the complexity and lends itself well to handling more complicated differentiations that involve logarithmic functions.

Exponential Functions

Exponential functions appear frequently in mathematics, often modeled by expressions like \(e^x\). Differentiating exponential functions involves recognizing their inherent simplicity, where the derivative of \(e^x\) remains \(e^x\). This unique property is a hallmark of exponential growth models.

However, exponential functions might also be presented in more complex forms such as \(e^{-x}\) or any variations where the exponent is a function of \(x\). In such cases, the Chain Rule often plays a role.

From the exercise, consider the term \(\frac{2}{e^x}\). Rewriting it as \(2e^{-x}\), we apply the Chain Rule. Differentiate \(e^{-x}\) as \(-e^{-x}\), thus yielding \(-2e^{-x}\) overall.

• Identify the exponential part and its exponent.
• Apply the Chain Rule if the exponent is a function of \(x\).
• Remember that the derivative of \(e^x\) is straightforward: \(e^x\).

Understanding these rules can greatly simplify the process of dealing with exponential expressions in calculus.