Answer :
Sure, I'll guide you through solving the two problems step-by-step.
### Problem 8: Finding [tex]\( \frac{dy}{dx} \)[/tex] using implicit differentiation
Given the equation:
[tex]\[ x^3 + 3x^2y + y^3 = 8 \][/tex]
1. Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^3 + 3x^2y + y^3) = \frac{d}{dx}(8) \][/tex]
The derivative of the right-hand side, a constant, is 0:
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
2. Apply the product rule and chain rule to the left-hand side:
- The derivative of [tex]\( x^3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 3x^2 \)[/tex].
- For [tex]\( 3x^2y \)[/tex], use the product rule:
[tex]\[ \frac{d}{dx}(3x^2y) = 3x^2\frac{dy}{dx} + 6xy \][/tex]
- The derivative of [tex]\( y^3 \)[/tex] with respect to [tex]\( x \)[/tex] using the chain rule is:
[tex]\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \][/tex]
3. Combine these derivatives:
[tex]\[ 3x^2 + 3x^2\frac{dy}{dx} + 6xy + 3y^2\frac{dy}{dx} = 0 \][/tex]
4. Collect the terms involving [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 3x^2 + 6xy + \left(3x^2 + 3y^2\right) \frac{dy}{dx} = 0 \][/tex]
5. Isolate [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 6xy + 3x^2 = - \left(3x^2 + 3y^2\right) \frac{dy}{dx} \][/tex]
[tex]\[ 6xy + 3x^2 = - 3(x^2 + y^2) \frac{dy}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{-(6xy + 3x^2)}{- 3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{6xy + 3x^2}{3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x^2 + 2xy}{x^2 + y^2} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\frac{x^2 + 2xy}{x^2 + y^2}} \][/tex]
Neither of the choices A or B match the calculation exactly, so double-checking and correcting one of the computations or choices might be necessary.
### Problem 9: Find the linearization [tex]\( L(x) \)[/tex] of [tex]\( f(x) \)[/tex] at [tex]\( x=a \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 + 5x + 5 \][/tex]
[tex]\[ a = 3 \][/tex]
1. Find [tex]\( f'(x) \)[/tex], the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(2x^2 + 5x + 5) \][/tex]
[tex]\[ f'(x) = 4x + 5 \][/tex]
2. Evaluate [tex]\( f(a) \)[/tex] and [tex]\( f'(a) \)[/tex]:
[tex]\[ f(3) = 2(3)^2 + 5(3) + 5 \][/tex]
[tex]\[ f(3) = 2 \cdot 9 + 15 + 5 \][/tex]
[tex]\[ f(3) = 18 + 15 + 5 \][/tex]
[tex]\[ f(3) = 38 \][/tex]
[tex]\[ f'(3) = 4(3) + 5 \][/tex]
[tex]\[ f'(3) = 12 + 5 \][/tex]
[tex]\[ f'(3) = 17 \][/tex]
3. Form the linearization [tex]\( L(x) \)[/tex]:
The linearization at [tex]\( x = a \)[/tex] is given by:
[tex]\[ L(x) = f(a) + f'(a)(x - a) \][/tex]
[tex]\[ L(x) = 38 + 17(x - 3) \][/tex]
4. Simplify [tex]\( L(x) \)[/tex]:
[tex]\[ L(x) = 38 + 17x - 51 \][/tex]
[tex]\[ L(x) = 17x - 13 \][/tex]
So, the linearization [tex]\( L(x) \)[/tex] at [tex]\( x = 3 \)[/tex] is:
[tex]\[ \boxed{L(x) = 38} \][/tex]
If should you not simplify it.
### Problem 8: Finding [tex]\( \frac{dy}{dx} \)[/tex] using implicit differentiation
Given the equation:
[tex]\[ x^3 + 3x^2y + y^3 = 8 \][/tex]
1. Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^3 + 3x^2y + y^3) = \frac{d}{dx}(8) \][/tex]
The derivative of the right-hand side, a constant, is 0:
[tex]\[ \frac{d}{dx}(8) = 0 \][/tex]
2. Apply the product rule and chain rule to the left-hand side:
- The derivative of [tex]\( x^3 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 3x^2 \)[/tex].
- For [tex]\( 3x^2y \)[/tex], use the product rule:
[tex]\[ \frac{d}{dx}(3x^2y) = 3x^2\frac{dy}{dx} + 6xy \][/tex]
- The derivative of [tex]\( y^3 \)[/tex] with respect to [tex]\( x \)[/tex] using the chain rule is:
[tex]\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \][/tex]
3. Combine these derivatives:
[tex]\[ 3x^2 + 3x^2\frac{dy}{dx} + 6xy + 3y^2\frac{dy}{dx} = 0 \][/tex]
4. Collect the terms involving [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 3x^2 + 6xy + \left(3x^2 + 3y^2\right) \frac{dy}{dx} = 0 \][/tex]
5. Isolate [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ 6xy + 3x^2 = - \left(3x^2 + 3y^2\right) \frac{dy}{dx} \][/tex]
[tex]\[ 6xy + 3x^2 = - 3(x^2 + y^2) \frac{dy}{dx} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{-(6xy + 3x^2)}{- 3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{6xy + 3x^2}{3(x^2 + y^2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x^2 + 2xy}{x^2 + y^2} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\frac{x^2 + 2xy}{x^2 + y^2}} \][/tex]
Neither of the choices A or B match the calculation exactly, so double-checking and correcting one of the computations or choices might be necessary.
### Problem 9: Find the linearization [tex]\( L(x) \)[/tex] of [tex]\( f(x) \)[/tex] at [tex]\( x=a \)[/tex]
Given:
[tex]\[ f(x) = 2x^2 + 5x + 5 \][/tex]
[tex]\[ a = 3 \][/tex]
1. Find [tex]\( f'(x) \)[/tex], the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(2x^2 + 5x + 5) \][/tex]
[tex]\[ f'(x) = 4x + 5 \][/tex]
2. Evaluate [tex]\( f(a) \)[/tex] and [tex]\( f'(a) \)[/tex]:
[tex]\[ f(3) = 2(3)^2 + 5(3) + 5 \][/tex]
[tex]\[ f(3) = 2 \cdot 9 + 15 + 5 \][/tex]
[tex]\[ f(3) = 18 + 15 + 5 \][/tex]
[tex]\[ f(3) = 38 \][/tex]
[tex]\[ f'(3) = 4(3) + 5 \][/tex]
[tex]\[ f'(3) = 12 + 5 \][/tex]
[tex]\[ f'(3) = 17 \][/tex]
3. Form the linearization [tex]\( L(x) \)[/tex]:
The linearization at [tex]\( x = a \)[/tex] is given by:
[tex]\[ L(x) = f(a) + f'(a)(x - a) \][/tex]
[tex]\[ L(x) = 38 + 17(x - 3) \][/tex]
4. Simplify [tex]\( L(x) \)[/tex]:
[tex]\[ L(x) = 38 + 17x - 51 \][/tex]
[tex]\[ L(x) = 17x - 13 \][/tex]
So, the linearization [tex]\( L(x) \)[/tex] at [tex]\( x = 3 \)[/tex] is:
[tex]\[ \boxed{L(x) = 38} \][/tex]
If should you not simplify it.
