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Sagot :
To find [tex]\( y' \)[/tex] and the slope of the tangent line to the graph of [tex]\( (3x - 4y)^5 = 2y^2 - 7 \)[/tex] at the point [tex]\( (3,2) \)[/tex], we follow these steps:
### 1. Differentiate Implicitly
Given the equation:
[tex]\[ (3x - 4y)^5 = 2y^2 - 7 \][/tex]
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left( (3x - 4y)^5 \right) = \frac{d}{dx} \left( 2y^2 - 7 \right) \][/tex]
Use the chain rule on the left-hand side:
[tex]\[ 5(3x - 4y)^4 \cdot \frac{d}{dx} (3x - 4y) \][/tex]
On the right-hand side, apply the chain rule as well:
[tex]\[ 4y \cdot \frac{dy}{dx} = 4yy' \][/tex]
Now, differentiate inside the parentheses [tex]\( 3x - 4y \)[/tex]:
[tex]\[ \frac{d}{dx} (3x - 4y) = 3 - 4\frac{dy}{dx} = 3 - 4y' \][/tex]
Putting it all together, we have:
[tex]\[ 5(3x - 4y)^4 \cdot (3 - 4y') = 4yy' \][/tex]
### 2. Solve for [tex]\( y' \)[/tex]
Rearrange to isolate [tex]\( y' \)[/tex]:
[tex]\[ 5(3x - 4y)^4 \cdot 3 - 20(3x - 4y)^4 y' = 4yy' \][/tex]
Combine like terms involving [tex]\( y' \)[/tex]:
[tex]\[ 15(3x - 4y)^4 = 20(3x - 4y)^4 y' + 4yy' \][/tex]
Factor out [tex]\( y' \)[/tex]:
[tex]\[ 15(3x - 4y)^4 = y'(20(3x - 4y)^4 + 4y) \][/tex]
Solve for [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{15(3x - 4y)^4}{20(3x - 4y)^4 + 4y} \][/tex]
Hence,
[tex]\[ y' = \frac{15(3x - 4y)^4}{4(y + 5(3x - 4y)^4)} \][/tex]
### 3. Evaluate [tex]\( y' \)[/tex] at the Point [tex]\( (3, 2) \)[/tex]
Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 2 \)[/tex] into the expression for [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{15(3 \cdot 3 - 4 \cdot 2)^4}{4(2 + 5(3 \cdot 3 - 4 \cdot 2)^4)} \][/tex]
Calculate the term [tex]\( 3x - 4y \)[/tex] at the given point:
[tex]\[ 3 \cdot 3 - 4 \cdot 2 = 9 - 8 = 1 \][/tex]
Substitute this back into the expression:
[tex]\[ y' = \frac{15 \cdot 1^4}{4 \left( 2 + 5 \cdot 1^4 \right)} \][/tex]
[tex]\[ y' = \frac{15}{4(2 + 5)} \][/tex]
[tex]\[ y' = \frac{15}{4 \cdot 7} \][/tex]
[tex]\[ y' = \frac{15}{28} \][/tex]
### Final Answer
[tex]\[ y' = \frac{15(3x - 4y)^4}{4 \left( y + 5(3x - 4y)^4 \right)} \][/tex]
At the point [tex]\( (3, 2) \)[/tex]:
[tex]\[ y' \bigg|_{(3,2)} = \frac{15}{28} \][/tex]
So, the derivatives are:
[tex]$ y' = \frac{15(3x - 4y)^4}{4 \left( y + 5(3x - 4y)^4 \right)} $[/tex]
and the slope of the tangent line at the point [tex]\( (3, 2) \)[/tex] is:
[tex]$ \left. y' \right|_{(3,2)} = \frac{15}{28} $[/tex]
### 1. Differentiate Implicitly
Given the equation:
[tex]\[ (3x - 4y)^5 = 2y^2 - 7 \][/tex]
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left( (3x - 4y)^5 \right) = \frac{d}{dx} \left( 2y^2 - 7 \right) \][/tex]
Use the chain rule on the left-hand side:
[tex]\[ 5(3x - 4y)^4 \cdot \frac{d}{dx} (3x - 4y) \][/tex]
On the right-hand side, apply the chain rule as well:
[tex]\[ 4y \cdot \frac{dy}{dx} = 4yy' \][/tex]
Now, differentiate inside the parentheses [tex]\( 3x - 4y \)[/tex]:
[tex]\[ \frac{d}{dx} (3x - 4y) = 3 - 4\frac{dy}{dx} = 3 - 4y' \][/tex]
Putting it all together, we have:
[tex]\[ 5(3x - 4y)^4 \cdot (3 - 4y') = 4yy' \][/tex]
### 2. Solve for [tex]\( y' \)[/tex]
Rearrange to isolate [tex]\( y' \)[/tex]:
[tex]\[ 5(3x - 4y)^4 \cdot 3 - 20(3x - 4y)^4 y' = 4yy' \][/tex]
Combine like terms involving [tex]\( y' \)[/tex]:
[tex]\[ 15(3x - 4y)^4 = 20(3x - 4y)^4 y' + 4yy' \][/tex]
Factor out [tex]\( y' \)[/tex]:
[tex]\[ 15(3x - 4y)^4 = y'(20(3x - 4y)^4 + 4y) \][/tex]
Solve for [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{15(3x - 4y)^4}{20(3x - 4y)^4 + 4y} \][/tex]
Hence,
[tex]\[ y' = \frac{15(3x - 4y)^4}{4(y + 5(3x - 4y)^4)} \][/tex]
### 3. Evaluate [tex]\( y' \)[/tex] at the Point [tex]\( (3, 2) \)[/tex]
Substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = 2 \)[/tex] into the expression for [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{15(3 \cdot 3 - 4 \cdot 2)^4}{4(2 + 5(3 \cdot 3 - 4 \cdot 2)^4)} \][/tex]
Calculate the term [tex]\( 3x - 4y \)[/tex] at the given point:
[tex]\[ 3 \cdot 3 - 4 \cdot 2 = 9 - 8 = 1 \][/tex]
Substitute this back into the expression:
[tex]\[ y' = \frac{15 \cdot 1^4}{4 \left( 2 + 5 \cdot 1^4 \right)} \][/tex]
[tex]\[ y' = \frac{15}{4(2 + 5)} \][/tex]
[tex]\[ y' = \frac{15}{4 \cdot 7} \][/tex]
[tex]\[ y' = \frac{15}{28} \][/tex]
### Final Answer
[tex]\[ y' = \frac{15(3x - 4y)^4}{4 \left( y + 5(3x - 4y)^4 \right)} \][/tex]
At the point [tex]\( (3, 2) \)[/tex]:
[tex]\[ y' \bigg|_{(3,2)} = \frac{15}{28} \][/tex]
So, the derivatives are:
[tex]$ y' = \frac{15(3x - 4y)^4}{4 \left( y + 5(3x - 4y)^4 \right)} $[/tex]
and the slope of the tangent line at the point [tex]\( (3, 2) \)[/tex] is:
[tex]$ \left. y' \right|_{(3,2)} = \frac{15}{28} $[/tex]
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