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Find \(f^{\prime}(x)\) $$ f(x)=4 \ln (5 x)-3 x^{-3}+4 x^{2} $$

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{4}{x} + 9x^{-4} + 8x \).

Step by step solution

01

Differentiate the Logarithmic Function

The function given is \( f(x) = 4 \ln(5x) - 3x^{-3} + 4x^2 \). Start with the differentiation of \( 4 \ln(5x) \). By the chain rule, the derivative of \( \ln(5x) \) is \( \frac{1}{5x} \times 5 = \frac{1}{x} \). Therefore, the derivative of \( 4 \ln(5x) \) is \( 4 \times \frac{1}{x} = \frac{4}{x} \).
02

Differentiate the Power Function with Negative Exponent

Next, differentiate \( -3x^{-3} \). Using the power rule \( \frac{d}{dx} x^n = nx^{n-1} \), the derivative is \( -3(-3)x^{-3-1} = 9x^{-4} \).
03

Differentiate the Quadratic Term

Differentiate the term \( 4x^2 \) using the power rule. The differentiation yields \( 2 \cdot 4 \cdot x^{2-1} = 8x \).
04

Combine All Derivatives

Finally, combine all the derivatives from steps 1 to 3 to form \( f'(x) \). Adding them together gives:\[ f'(x) = \frac{4}{x} + 9x^{-4} + 8x \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule for Derivatives
The power rule is one of the most essential tools for finding derivatives. It's simple and efficient for differentiating functions where terms involve variables raised to a power. The power rule states that if you have a term in the form of \( x^n \), its derivative is \( nx^{n-1} \). This means you multiply the exponent by the coefficient and then subtract one from the exponent.
  • For example, in the term \( -3x^{-3} \), apply the power rule: multiply \(-3\) (the coefficient) by the exponent \(-3\), resulting in \( 9x^{-4} \) after decreasing the power by one.
  • Similarly, for \( 4x^2 \), multiply \( 4 \) by \( 2 \) and reduce the exponent to get \( 8x \).
The power rule is versatile and comes in handy when working with polynomial expressions, making it a must-know concept in calculus. It simplifies the process by breaking down each term based on its individual power, enabling quick and correct differentiation.
Chain Rule for Composite Functions
The chain rule is crucial when dealing with composite functions—functions within other functions. It allows you to differentiate complex expressions by understanding how to "break them apart" effectively. The chain rule can be remembered as:
  • If you have a function \(f(g(x))\), then its derivative \(f'(g(x))\) is \(f'(g(x)) \cdot g'(x)\).
For example, consider \(4 \ln(5x)\), a composite function because \(5x\) is inside the logarithmic function. Here, think of \(5x\) as \(g(x)\) and \(\ln(u)\) as \(f(u)\). Applying chain rule, you first find the derivative of the inside function \(5x\) which is 5, then multiply by the derivative of the outer function \(\ln(u)\), where the derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). Thus, this computes as \( \frac{1}{x} \) because of the substitution back \(u=5x\), finally equaling \( \frac{4}{x} \) when multiplied by 4.
Logarithmic Function Derivative
The derivative of a logarithmic function is a key concept that is often encountered. When you differentiate \( \ln(x) \), the result is \( \frac{1}{x} \). This is because the natural logarithm function is the inverse of the exponential function, and this unique inverse relationship is reflected in the simple form of its derivative.
  • When differentiating a function like \( \ln(5x) \), you need to recognize the inner function, which is \(5x\).
  • According to the chain rule, you calculate \( \frac{1}{5x} \) and multiply it by the derivative of \(5x\), which is 5, resulting in \( \frac{1}{x} \).
Logarithmic differentiation is particularly useful when dealing with products, quotients, or powers of variables. Understanding how to apply these rules allows for seamless differentiation of complex functions involving logarithms, often resulting in much simpler expressions.

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Most popular questions from this chapter

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Graph the function \(y=f(x)=x^{5}+3 x^{2}-9 x+4\) on your graphing calculator. a. Roughly estimate the values where the tangent to the graph of \(y=f(x)\) is horizontal. b. Now graph \(y=f^{\prime}(x)\) and solve \(f^{\prime}(x)=0\) using your computer or graphing calculator. c. How do you relate the solutions of \(f^{\prime}(x)=0\) found in part (b) to the values of \(x\) found in part (a)? d. Can you confirm your answers found in part (b) using calculus? Why or why not?

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