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To find the derivative of the function [tex]\( y = \frac{x-1}{2x^2 - 7x + 5} \)[/tex] with respect to [tex]\( x \)[/tex] and then evaluate this derivative at [tex]\( x = 2 \)[/tex], we can follow these steps:
1. Identify the function and its components:
Given function:
[tex]\[ y = \frac{x-1}{2x^2 - 7x + 5} \][/tex]
2. Apply the quotient rule:
The quotient rule for differentiation states:
[tex]\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \][/tex]
where [tex]\( u = x - 1 \)[/tex] and [tex]\( v = 2x^2 - 7x + 5 \)[/tex].
3. Compute [tex]\( u' \)[/tex] and [tex]\( v' \)[/tex]:
[tex]\[ u = x - 1 \implies u' = 1 \][/tex]
[tex]\[ v = 2x^2 - 7x + 5 \implies v' = 4x - 7 \][/tex]
4. Apply the quotient rule with these components:
[tex]\[ \frac{dy}{dx} = \frac{(1)(2x^2 - 7x + 5) - (x-1)(4x - 7)}{(2x^2 - 7x + 5)^2} \][/tex]
5. Simplify the expression:
Expand the numerator:
[tex]\[ \text{Numerator} = (2x^2 - 7x + 5) - (x-1)(4x - 7) \][/tex]
Expand [tex]\( (x-1)(4x - 7) \)[/tex]:
[tex]\[ (x-1)(4x - 7) = 4x^2 - 7x - 4x + 7 = 4x^2 - 11x + 7 \][/tex]
Subtract this from [tex]\( (2x^2 - 7x + 5) \)[/tex]:
[tex]\[ \text{Numerator} = 2x^2 - 7x + 5 - (4x^2 - 11x + 7) \][/tex]
[tex]\[ = 2x^2 - 7x + 5 - 4x^2 + 11x - 7 \][/tex]
[tex]\[ = -2x^2 + 4x - 2 \][/tex]
6. Fit it back into the derivative expression:
[tex]\[ \frac{dy}{dx} = \frac{-2x^2 + 4x - 2}{(2x^2 - 7x + 5)^2} \][/tex]
7. Evaluate at [tex]\( x = 2 \)[/tex]:
Substituting [tex]\( x = 2 \)[/tex] into the derivative:
[tex]\[ \frac{dy}{dx} \bigg|_{x=2} = \frac{-2(2)^2 + 4(2) - 2}{(2(2)^2 - 7(2) + 5)^2} \][/tex]
[tex]\[ = \frac{-2(4) + 8 - 2}{(8 - 14 + 5)^2} \][/tex]
[tex]\[ = \frac{-8 + 8 - 2}{(-1)^2} \][/tex]
[tex]\[ = \frac{-2}{1} \][/tex]
[tex]\[ = -2 \][/tex]
Therefore, the derivative [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( x = 2 \)[/tex] is [tex]\( -2 \)[/tex].
1. Identify the function and its components:
Given function:
[tex]\[ y = \frac{x-1}{2x^2 - 7x + 5} \][/tex]
2. Apply the quotient rule:
The quotient rule for differentiation states:
[tex]\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \][/tex]
where [tex]\( u = x - 1 \)[/tex] and [tex]\( v = 2x^2 - 7x + 5 \)[/tex].
3. Compute [tex]\( u' \)[/tex] and [tex]\( v' \)[/tex]:
[tex]\[ u = x - 1 \implies u' = 1 \][/tex]
[tex]\[ v = 2x^2 - 7x + 5 \implies v' = 4x - 7 \][/tex]
4. Apply the quotient rule with these components:
[tex]\[ \frac{dy}{dx} = \frac{(1)(2x^2 - 7x + 5) - (x-1)(4x - 7)}{(2x^2 - 7x + 5)^2} \][/tex]
5. Simplify the expression:
Expand the numerator:
[tex]\[ \text{Numerator} = (2x^2 - 7x + 5) - (x-1)(4x - 7) \][/tex]
Expand [tex]\( (x-1)(4x - 7) \)[/tex]:
[tex]\[ (x-1)(4x - 7) = 4x^2 - 7x - 4x + 7 = 4x^2 - 11x + 7 \][/tex]
Subtract this from [tex]\( (2x^2 - 7x + 5) \)[/tex]:
[tex]\[ \text{Numerator} = 2x^2 - 7x + 5 - (4x^2 - 11x + 7) \][/tex]
[tex]\[ = 2x^2 - 7x + 5 - 4x^2 + 11x - 7 \][/tex]
[tex]\[ = -2x^2 + 4x - 2 \][/tex]
6. Fit it back into the derivative expression:
[tex]\[ \frac{dy}{dx} = \frac{-2x^2 + 4x - 2}{(2x^2 - 7x + 5)^2} \][/tex]
7. Evaluate at [tex]\( x = 2 \)[/tex]:
Substituting [tex]\( x = 2 \)[/tex] into the derivative:
[tex]\[ \frac{dy}{dx} \bigg|_{x=2} = \frac{-2(2)^2 + 4(2) - 2}{(2(2)^2 - 7(2) + 5)^2} \][/tex]
[tex]\[ = \frac{-2(4) + 8 - 2}{(8 - 14 + 5)^2} \][/tex]
[tex]\[ = \frac{-8 + 8 - 2}{(-1)^2} \][/tex]
[tex]\[ = \frac{-2}{1} \][/tex]
[tex]\[ = -2 \][/tex]
Therefore, the derivative [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( x = 2 \)[/tex] is [tex]\( -2 \)[/tex].
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