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Find the derivative of the following functions. $$f(x)=\frac{x^{3}-4 x^{2}+x}{x-2}$$

Short Answer

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Question: Determine the derivative of the function $$f(x) = \frac{x^{3} - 4x^{2} + x}{x - 2}$$. Answer: The derivative of the function f(x) is $$f'(x) = \frac{-x^{2} + 12x - 6}{(x-2)^{2}}$$.

Step by step solution

01

Find the derivative of u(x)

To find the derivative of u(x), we will differentiate each term with respect to x: $$u'(x) = \frac{d}{dx}(x^{3} - 4x^{2} + x)$$ Using the power rule for differentiation, we get: $$u'(x) = 3x^{2} - 8x + 1$$
02

Find the derivative of v(x)

To find the derivative of v(x), we will differentiate it with respect to x: $$v'(x) = \frac{d}{dx}(x - 2)$$ Using the power rule for differentiation, we get: $$v'(x) = 1$$
03

Apply the quotient rule

Now we have all the pieces we need to apply the quotient rule. Plugging u(x), u'(x), v(x), and v'(x) into the quotient rule formula, we get: $$f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^{2}}$$ Substitute the expressions for u(x), u'(x), v(x), and v'(x): $$f'(x) = \frac{(x-2)(3x^{2} - 8x + 1) - (x^{3} - 4x^{2} + x)(1)}{(x-2)^{2}}$$
04

Simplify the expression

Now let's simplify the expression for f'(x): $$f'(x) = \frac{3x^{3} - 8x^{2} + x - 6x^{2} + 16x - 6 - x^{3} + 4x^{2} - x}{(x-2)^{2}}$$ Combine like terms: $$f'(x) = \frac{-x^{2} + 12x - 6}{(x-2)^{2}}$$ Now, the derivative of the given function f(x) is: $$f'(x) = \frac{-x^{2} + 12x - 6}{(x-2)^{2}}$$

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Key Concepts

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

Quotient Rule for Differentiation
When faced with the task of finding the derivative of a rational function, which is a ratio of two functions, we utilize the quotient rule for differentiation. The quotient rule is particularly useful when dividing one polynomial by another.

The quotient rule states that if we have a function in the form of \(f(x) = \frac{u(x)}{v(x)}\), where both \(u\) and \(v\) are differentiable functions, the derivative of \(f\) with respect to \(x\) is:
\[f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\]
This formula essentially tells us to differentiate the top function (nominator), multiply it by the bottom function (denominator), then subtract the product of the top function and the derivative of the bottom function. All this is divided by the square of the bottom function. When applying the quotient rule, it's crucial to keep track of each term and perform the operations accurately to avoid errors.
Power Rule for Differentiation
The power rule simplifies the process of taking the derivative of monomials, which are individual terms consisting of a coefficient multiplied by a variable raised to a power. According to the power rule, if \(f(x) = x^n\) where \(n\) is any real number, the derivative of \(f\) is:
\[f'(x) = nx^{n-1}\]
This rule is beautifully straightforward: bring down the exponent as a coefficient, and subtract one from the exponent. In the step by step solution, the power rule enabled us to find \(u'(x)\) efficiently. This technique is a foundational tool in calculus that allows students to quickly find derivatives without complex calculations.
Simplifying Algebraic Expressions
Simplifying algebraic expressions is a vital skill that helps in reducing complex forms into simpler ones, making it easier to understand or further manipulate the expression. After applying the quotient and power rules for differentiation, simplifying involves combining like terms, factoring, and canceling terms when appropriate.

In the provided example, we simplify the derivative \(f'(x)\) by expanding the products and then combining the terms with the same powers of \(x\). The goal is to end up with the most straightforward expression for the derivative, as seen in the step by step solution where we obtained \(f'(x) = \frac{-x^2 + 12x - 6}{(x-2)^2}\). Simplification renders the derivative more comprehensible and, where applicable, more suitable for further analysis or graphing.
Differentiation Techniques
Differentiation techniques are an assortment of methods and rules utilized to find the derivative of various functions. Each technique corresponds to a different type of function or operation, such as products, quotients, or compositions of functions.

In calculus, mastering various differentiation techniques, like the quotient and power rules, chain rule, and product rule, among others, enables students to differentiate a broad range of functions. These techniques are not standalone; often, you need to combine them to find the derivative of more complex functions. For instance, in the provided exercise, we first used the power rule to find \(u'(x)\) and \(v'(x)\), and then we employed the quotient rule to differentiate the ratio \(\frac{u(x)}{v(x)}\). Constant practice with these techniques will lead to more confidence and efficiency in tackling calculus problems.

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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 \((c)\). $$C(x)=-0.02 x^{2}+50 x+100, p(x)=100, a=500$$

A 500-liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min} .\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank (in liters) is given by \(V(t)=500-0.5 t\). a. Graph the mass function and verify that \(M(0)=0\). b. Graph the volume function and verify that the tank is empty when \(t=1000\) min. c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) and \(\lim _{\theta \rightarrow 000^{-}} C(t) ?\) \(t \rightarrow 1\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\). e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\). f. For what times is the concentration of the solution increasing? Decreasing?

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(2 x)^{2 x}$$.

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{10 x}\right)$$

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