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Use logarithmic differentiation to evaluate $f^{\prime}(x)$$. $$f(x)=\frac{x^{8} \cos ^{3} x}{\sqrt{x-1}}$$

Short Answer

Expert verified
Question: Find the derivative of the function $$f(x) = \frac{x^{8}\cos^{3}{x}}{\sqrt{x-1}}$$ using logarithmic differentiation. Answer: The derivative of the function f(x) using logarithmic differentiation is $$f'(x) = \frac{8x^{7}\cos^{3}{x} - 3x^{8}\cos^{2}{x}\sin{x} - \frac{1}{2}x^{8}\cos^{3}{x}}{2(x-1)\sqrt{x-1}}$$.

Step by step solution

01

Logarithm of both sides

Start by taking the natural logarithm of both sides of the equation: $$\ln(f(x)) = \ln\left(\frac{x^{8} \cos^{3}{x}}{\sqrt{x-1}}\right)$$
02

Simplify using logarithm properties

Using logarithm properties for products and quotients, we can simplify the equation as follows: $$\ln(f(x)) = 8\ln{x} + 3\ln{(\cos{x})} - \frac{1}{2}\ln{(x-1)}$$
03

Differentiate both sides with respect to x

Now, we'll differentiate each term with respect to x using the chain rule and standard derivatives for ln(x) and ln(cos(x)): $$\frac{f'(x)}{f(x)} = 8\frac{1}{x} - 3\frac{\sin{x}}{\cos{x}} - \frac{1}{2}\frac{1}{x-1}$$
04

Isolate f'(x)

Now, we need to isolate f'(x) in the equation, which can be done by multiplying both sides by f(x): $$f'(x) = f(x) \cdot \left(8\frac{1}{x} - 3\frac{\sin{x}}{\cos{x}} - \frac{1}{2}\frac{1}{x-1}\right)$$ Now insert the original expression for f(x): $$f'(x) = \frac{x^{8} \cos ^{3} x}{\sqrt{x-1}} \cdot \left(8\frac{1}{x} - 3\frac{\sin{x}}{\cos{x}} - \frac{1}{2}\frac{1}{x-1}\right)$$
05

Simplify the expression

Simplify the expression for f'(x) by canceling out common terms and applying algebraic manipulations: $$f'(x) = \frac{8x^{7}\cos^{3}{x} - 3x^{8}\cos^{2}{x}\sin{x} - \frac{1}{2}x^{8}\cos^{3}{x}}{2(x-1)\sqrt{x-1}}$$ So the derivative of the function f(x) using logarithmic differentiation is: $$f'(x) = \frac{8x^{7}\cos^{3}{x} - 3x^{8}\cos^{2}{x}\sin{x} - \frac{1}{2}x^{8}\cos^{3}{x}}{2(x-1)\sqrt{x-1}}$$

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. In its essence, the chain rule provides a way to differentiate a function that is made up of one function inside another. For instance, if you have a function \(h(x) = g(f(x))\), the derivative \(h'(x)\) will be \(g'(f(x)) \times f'(x)\).

Here's why this is important in logarithmic differentiation. When taking the derivative of a logarithmic function that involves another function—as we see with \(\frac{d}{dx}\big(\text{ln}(f(x))\big)\)—the chain rule allows us to differentiate the outer function, which is the natural logarithm, and multiply it by the derivative of the inner function, \(f(x)\). This is precisely what we utilize when applying the derivative to \(\text{ln}(x)\) or \(\text{ln}(\text{cos}(x))\) in the given exercise. The chain rule ensures that every layer of the function is accounted for in the differentiation process.
Natural Logarithm Properties
Understanding the properties of the natural logarithm is crucial when simplifying expressions before differentiation. The natural logarithm, denoted as \(\text{ln}(x)\), has several key properties that make logarithmic differentiation handy. These properties include:
  • The product rule: \(\text{ln}(xy) = \text{ln}(x) + \text{ln}(y)\).
  • The quotient rule: \(\text{ln}\big(\frac{x}{y}\big) = \text{ln}(x) - \text{ln}(y)\).
  • The power rule: \(\text{ln}(x^a) = a \text{ln}(x)\).

During the step-by-stepprocess, we visit the second step wherein we simplify the complex function using these properties. The original function \(f(x)\) includes a product, a quotient, and an exponent within its structure. By applying natural logarithm properties, the equation was significantly simplified and prepared for the application of the derivative. These properties allowed us to break down the composite logarithm into the sum and difference of several simpler logarithms, each of which is easier to differentiate.
Derivative of Logarithmic Functions
When dealing with the derivative of logarithmic functions, there are straightforward rules to follow. For example, the derivative of the natural logarithm of x, denoted as \(\text{ln}(x)\), with respect to x is \(\frac{d}{dx}(\text{ln}(x)) = \frac{1}{x}\). Furthermore, the differentiation of more complex logarithmic functions often involves utilizing the chain rule, as discussed earlier.

In our exercise, step three is where these rules come into play. After simplifying the logarithmic expression in the previous step, we differentiate both sides of the equation. The derivatives of \(\text{ln}(x)\) and \(\text{ln}(\text{cos}(x))\) are found using the aforementioned rule and the chain rule, producing a term-by-term differentiation that leads to \(\frac{f'(x)}{f(x)}\). This relationship is crucial as it allows us to isolate \(f'(x)\) by multiplying both sides by \(f(x)\) in the subsequent step. The exercise exemplifies how the derivative of each logarithmic part is determined and then combined to find the overall derivative of the original function.

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Most popular questions from this chapter

Let \(C(x)\) represent the cost of producing \(x\) items and \(p(x)\) be the sale price per item if \(x\) items are sold. The profit \(P(x)\) of selling x items is \(P(x)=x p(x)-C(x)\) (revenue minus costs). The average profit per item when \(x\) items are sold is \(P(x) / x\) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that \(x\) items have already been sold. Consider the following cost functions \(C\) and price functions \(p\). a. Find the profit function \(P\). b. Find the average profit function and marginal profit function. c. Find the average profit and marginal profit if \(x=a\) units are sold. d. Interpret the meaning of the values obtained in part \((c)\). $$C(x)=-0.02 x^{2}+50 x+100, p(x)=100, a=500$$

The position (in meters) of a marble rolling up a long incline is given by \(s=\frac{100 t}{t+1},\) where \(t\) is measured in seconds and \(s=0\) is the starting point. a. Graph the position function. b. Find the velocity function for the marble. c. Graph the velocity function and give a description of the motion of the marble. d. At what time is the marble 80 m from its starting point? e. At what time is the velocity \(50 \mathrm{m} / \mathrm{s} ?\)

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(2 \sin ^{-1} x\right)=1-2 x^{2}$$

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\log _{2} \frac{8}{\sqrt{x+1}}$$

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