Questions: 2.9 The Quotient Rule Calculus Name: 1. h(x) = (r^2 + tr - 1) / (x - 1) h'(x) = L(2x + 4)(x - 3)^-3 - (x^2(4x - 1) + 1) 2(x^2) f'(x) = (2x^2 - 6x + 4x - 12 - x^2 - 4x + 1) / (x - 3)^2 (x^2 - 6x - 11) / (x - 3)^2 2. g(x) = set / (2 sin x) 3. f(x) = enas / (z - 16) 4. g(x) = (4x^3 - 5x^2 + 5x) / x^2 5. h(x) = 5x^2 / ln x (4x^3 - 5x^2 + 5x) = 12x^2 - 10x + 5 g'(x) = (12x^2 - 10x + 5)(x^2) - (4x^3 - 5x^2 + 5x)(2x) / (x^2)^2 g'(x) = (4x^4 - 5x^2) / x^4 g'(x) = 4 - 5/x^2 6. f(x) = (x + 2) / (x^2 + 2) 7. h(x) = (8x^3 + 4x^2 - 9x) / 2x

2.9 The Quotient Rule
Calculus
Name:
1. h(x) = (r^2 + tr - 1) / (x - 1)
h'(x) = L(2x + 4)(x - 3)^-3 - (x^2(4x - 1) + 1) 2(x^2)
f'(x) = (2x^2 - 6x + 4x - 12 - x^2 - 4x + 1) / (x - 3)^2
(x^2 - 6x - 11) / (x - 3)^2
2. g(x) = set / (2 sin x)
3. f(x) = enas / (z - 16)
4. g(x) = (4x^3 - 5x^2 + 5x) / x^2
5. h(x) = 5x^2 / ln x
(4x^3 - 5x^2 + 5x) = 12x^2 - 10x + 5
g'(x) = (12x^2 - 10x + 5)(x^2) - (4x^3 - 5x^2 + 5x)(2x) / (x^2)^2
g'(x) = (4x^4 - 5x^2) / x^4
g'(x) = 4 - 5/x^2
6. f(x) = (x + 2) / (x^2 + 2)
7. h(x) = (8x^3 + 4x^2 - 9x) / 2x
Transcript text: 2.9 The Quotient Rule Calcalus Name: 1. $h(x)=\frac{r^{2}+t r-1}{x-1}$ \[ \begin{array}{l} h^{\prime}(x)=L(2 x+4)(x-3)^{-3}-\left(x^{2}(4 x-1)+1\right) 2\left(x^{2}\right. \\ f^{\prime}(x)=\left(2 x^{2}-6 x+4 x-12-x^{2}-4 x+1\right) /\left(x-3^{2}=\right. \\ \frac{\left(x^{2}-6 x-11\right.}{(x-3)^{2}} \end{array} \] 2. $g(x)=\frac{\operatorname{set}}{2 \sin x}$ 3. $f(x)=\frac{e n a s}{z-16}$ 4. $g(x)=\frac{4 x^{3}-5 x^{2}+5 x}{x^{2}}$ 5. $h(x)=\frac{5 x^{2}}{\ln x}$ \[ \begin{array}{l} \left(4 x^{3}-5 x^{2}+5 x\right)=12 x^{2}-10 x+5 \\ g^{\prime}(x)=\frac{\left.12 x^{2}-10 x+5\right)\left(x^{2}\right)-\left(4 x^{3}-5 x^{2}+5 x\right)(2 x)}{\left(x^{2}\right)^{2}} \\ g^{\prime}(x)=\frac{4 x^{4}-5 x^{2}}{x^{4}} \\ g^{\prime}(x)=4-\frac{5}{x^{2}} \end{array} \] 6. $f(x)=\frac{x+2}{x^{2}+2}$ 7. $h(x)=\frac{8 x^{3}+4 x^{2}-9 x}{2 x}$
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Solution

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Solution Steps

Step 1: Identify the given function and apply the quotient rule

For the first problem, the function is \( h(x) = \frac{2x^2 - 3x + 1}{x - 1} \).

The quotient rule states that for a function \( h(x) = \frac{f(x)}{g(x)} \), the derivative \( h'(x) \) is given by: \[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

Here, \( f(x) = 2x^2 - 3x + 1 \) and \( g(x) = x - 1 \).

Step 2: Compute the derivatives of the numerator and the denominator

Compute \( f'(x) \) and \( g'(x) \): \[ f'(x) = \frac{d}{dx}(2x^2 - 3x + 1) = 4x - 3 \] \[ g'(x) = \frac{d}{dx}(x - 1) = 1 \]

Step 3: Apply the quotient rule formula

Substitute \( f(x) \), \( f'(x) \), \( g(x) \), and \( g'(x) \) into the quotient rule formula: \[ h'(x) = \frac{(4x - 3)(x - 1) - (2x^2 - 3x + 1)(1)}{(x - 1)^2} \]

Step 4: Simplify the expression

Expand and simplify the numerator: \[ (4x - 3)(x - 1) = 4x^2 - 4x - 3x + 3 = 4x^2 - 7x + 3 \] \[ (2x^2 - 3x + 1)(1) = 2x^2 - 3x + 1 \]

Combine the results: \[ h'(x) = \frac{4x^2 - 7x + 3 - (2x^2 - 3x + 1)}{(x - 1)^2} = \frac{4x^2 - 7x + 3 - 2x^2 + 3x - 1}{(x - 1)^2} = \frac{2x^2 - 4x + 2}{(x - 1)^2} \]

Final Answer

\[ h'(x) = \frac{2(x^2 - 2x + 1)}{(x - 1)^2} = \frac{2(x - 1)^2}{(x - 1)^2} = 2 \]

The derivative of the function \( h(x) = \frac{2x^2 - 3x + 1}{x - 1} \) is \( h'(x) = 2 \).

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