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Chapter 3: Problem 20

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

Short Answer

Expert verified
Question: Find the derivative of the function $$f(x) = \frac{x^3 - 4x^2 + x}{x - 2}$$. Answer: The derivative of the function is $$f'(x) = \frac{2x^3 - 10x^2 + 16x - 2}{(x-2)^2}$$.

Step by step solution

01

Identify u(x) and v(x)

The given function is $$f(x)=\frac{x^{3}-4 x^{2}+x}{x-2}$$. In this case, we have: - Numerator, u(x) = $$x^3 - 4x^2 + x$$ - Denominator, v(x) = $$x - 2$$
02

Calculate u'(x) and v'(x)

Now, we'll find the derivatives of u(x) and v(x) with respect to x: - $$u'(x) = \frac{d}{dx}(x^3 - 4x^2 + x) = 3x^2 - 8x + 1$$ - $$v'(x) = \frac{d}{dx}(x-2) = 1$$
03

Apply the Quotient Rule

Now that we have the derivatives of u(x) and v(x), we'll apply the quotient rule to find the derivative of the given function: $$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$$ So, $$f'(x) = \frac{(3x^2 - 8x + 1)(x -2) - (x^3 -4x^2 + x)(1)}{(x-2)^2}$$
04

Simplify the expression

Now we'll simplify the expression for f'(x): $$f'(x) = \frac{3x^3 - 8x^2 + x - 6x^2 + 16x -2 - x^3 + 4x^2 - x}{(x-2)^2}$$ Combine like terms: $$f'(x) = \frac{3x^3 -14x^2 + 17x - 2 - x^3 +4x^2 - x}{(x-2)^2}$$ $$f'(x) = \frac{2x^3 - 10x^2 + 16x - 2}{(x-2)^2}$$ So, the derivative of the given function is: $$f'(x) = \frac{2x^3 - 10x^2 + 16x - 2}{(x-2)^2}$$

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Most popular questions from this chapter

Special Product Rule In general, the derivative of a product is not the product of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f g\) equals \(f^{\prime} g^{\prime}\).

Use any method to evaluate the derivative of the following functions. $$h(x)=\left(5 x^{7}+5 x\right)\left(6 x^{3}+3 x^{2}+3\right)$$

Electrostatic force The magnitude of the electrostatic force between two point charges \(Q\) and \(q\) of the same sign is given by \(F(x)=\frac{k Q q}{x^{2}},\) where \(x\) is the distance (measured in meters) between the charges and \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\) is a physical constant (C stands for coulomb, the unit of charge; N stands for newton, the unit of force). a. Find the instantaneous rate of change of the force with respect to the distance between the charges. b. For two identical charges with \(Q=q=1 \mathrm{C},\) what is the instantaneous rate of change of the force at a separation of \(x=0.001 \mathrm{m} ?\) c. Does the magnitude of the instantaneous rate of change of the force increase or decrease with the separation? Explain.

Proof of the Quotient Rule Let \(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that \(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}\) b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=x^{3}+3$$
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