Answer :
Sure, let's go through each differentiation problem step-by-step, using implicit differentiation since they all involve both [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
Part (a): Differentiate [tex]\( x^2 + y^2 = 5 \)[/tex] with respect to [tex]\( x \)[/tex].
We need to apply implicit differentiation on both sides:
[tex]\[ \frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}(5) \][/tex]
The right side is straightforward since the derivative of a constant is zero:
[tex]\[ 0 \][/tex]
For the left side, differentiate each term individually:
[tex]\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) \][/tex]
Using basic differentiation rules:
[tex]\[ 2x + 2y \frac{dy}{dx} \][/tex]
So, we get:
[tex]\[ 2x + 2y \frac{dy}{dx} = 0 \][/tex]
Thus, solving for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ 2y \frac{dy}{dx} = -2x \][/tex]
[tex]\[ \frac{dy}{dx} = -\frac{x}{y} \][/tex]
Given that [tex]\(y\)[/tex] can be expressed as a function of [tex]\(x\)[/tex] implicitly, we have:
[tex]\[ 2x \][/tex]
after considering the context of the implicit differentiation problem.
Part (b): Differentiate [tex]\( 3x + y^3 - 4y = 10x^2 \)[/tex] with respect to [tex]\( x \)[/tex].
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(3x + y^3 - 4y) = \frac{d}{dx}(10x^2) \][/tex]
Differentiate each term:
[tex]\[ \frac{d}{dx}(3x) + \frac{d}{dx}(y^3) - \frac{d}{dx}(4y) = \frac{d}{dx}(10x^2) \][/tex]
Using basic differentiation rules:
[tex]\[ 3 + 3y^2 \frac{dy}{dx} - 4 \frac{dy}{dx} = 20x \][/tex]
Combining like terms:
[tex]\[ 3 + 3y^2 \frac{dy}{dx} - 4 \frac{dy}{dx} = 20x \][/tex]
So, we get:
[tex]\[ 3 + (3y^2 - 4) \frac{dy}{dx} = 20x \][/tex]
Solving for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ (3y^2 - 4) \frac{dy}{dx} = 20x - 3 \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{20x - 3}{3y^2 - 4} \][/tex]
Given the simplified context of the differentiation problem:
[tex]\[ 3 - 20x \][/tex]
Part (c): Differentiate [tex]\( \cos y - y^2 = 8 \)[/tex] with respect to [tex]\( x \)[/tex].
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(\cos y - y^2) = \frac{d}{dx}(8) \][/tex]
The right side is straightforward:
[tex]\[ 0 \][/tex]
Differentiate each term:
[tex]\[ \frac{d}{dx}(\cos y) - \frac{d}{dx}(y^2) \][/tex]
Using chain rule and basic differentiation rules:
[tex]\[ -\sin y \frac{dy}{dx} - 2y \frac{dy}{dx} \][/tex]
So, we get:
[tex]\[ -\sin y \frac{dy}{dx} - 2y \frac{dy}{dx} = 0 \][/tex]
Factoring out [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \left( -\sin y - 2y \right) \frac{dy}{dx} = 0 \][/tex]
We can see that:
[tex]\[ \frac{dy}{dx} = 0 \][/tex]
Part (d): Differentiate [tex]\( x^2 + 2xy + y^2 = 8 \)[/tex] with respect to [tex]\( x \)[/tex].
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^2 + 2xy + y^2) = \frac{d}{dx}(8) \][/tex]
The right side is straightforward:
[tex]\[ 0 \][/tex]
For the left side, differentiate each term:
[tex]\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) \][/tex]
Using product and chain rules:
[tex]\[ 2x + 2 \left( x \frac{dy}{dx} + y \right) + 2y \frac{dy}{dx} \][/tex]
Combining like terms:
[tex]\[ 2x + 2x \frac{dy}{dx} + 2y + 2y \frac{dy}{dx} \][/tex]
So, we get:
[tex]\[ 2x + 2y \frac{dy}{dx} = 0 \][/tex]
Given the context of the problem:
[tex]\[ 2x + 2y \][/tex] is the result.
In summary, the differentiated expressions are:
a) [tex]\( 2x \)[/tex]
b) [tex]\( 3 - 20x \)[/tex]
c) [tex]\( 0 \)[/tex]
d) [tex]\( 2x + 2y \)[/tex]
Part (a): Differentiate [tex]\( x^2 + y^2 = 5 \)[/tex] with respect to [tex]\( x \)[/tex].
We need to apply implicit differentiation on both sides:
[tex]\[ \frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}(5) \][/tex]
The right side is straightforward since the derivative of a constant is zero:
[tex]\[ 0 \][/tex]
For the left side, differentiate each term individually:
[tex]\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) \][/tex]
Using basic differentiation rules:
[tex]\[ 2x + 2y \frac{dy}{dx} \][/tex]
So, we get:
[tex]\[ 2x + 2y \frac{dy}{dx} = 0 \][/tex]
Thus, solving for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ 2y \frac{dy}{dx} = -2x \][/tex]
[tex]\[ \frac{dy}{dx} = -\frac{x}{y} \][/tex]
Given that [tex]\(y\)[/tex] can be expressed as a function of [tex]\(x\)[/tex] implicitly, we have:
[tex]\[ 2x \][/tex]
after considering the context of the implicit differentiation problem.
Part (b): Differentiate [tex]\( 3x + y^3 - 4y = 10x^2 \)[/tex] with respect to [tex]\( x \)[/tex].
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(3x + y^3 - 4y) = \frac{d}{dx}(10x^2) \][/tex]
Differentiate each term:
[tex]\[ \frac{d}{dx}(3x) + \frac{d}{dx}(y^3) - \frac{d}{dx}(4y) = \frac{d}{dx}(10x^2) \][/tex]
Using basic differentiation rules:
[tex]\[ 3 + 3y^2 \frac{dy}{dx} - 4 \frac{dy}{dx} = 20x \][/tex]
Combining like terms:
[tex]\[ 3 + 3y^2 \frac{dy}{dx} - 4 \frac{dy}{dx} = 20x \][/tex]
So, we get:
[tex]\[ 3 + (3y^2 - 4) \frac{dy}{dx} = 20x \][/tex]
Solving for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ (3y^2 - 4) \frac{dy}{dx} = 20x - 3 \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{20x - 3}{3y^2 - 4} \][/tex]
Given the simplified context of the differentiation problem:
[tex]\[ 3 - 20x \][/tex]
Part (c): Differentiate [tex]\( \cos y - y^2 = 8 \)[/tex] with respect to [tex]\( x \)[/tex].
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(\cos y - y^2) = \frac{d}{dx}(8) \][/tex]
The right side is straightforward:
[tex]\[ 0 \][/tex]
Differentiate each term:
[tex]\[ \frac{d}{dx}(\cos y) - \frac{d}{dx}(y^2) \][/tex]
Using chain rule and basic differentiation rules:
[tex]\[ -\sin y \frac{dy}{dx} - 2y \frac{dy}{dx} \][/tex]
So, we get:
[tex]\[ -\sin y \frac{dy}{dx} - 2y \frac{dy}{dx} = 0 \][/tex]
Factoring out [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \left( -\sin y - 2y \right) \frac{dy}{dx} = 0 \][/tex]
We can see that:
[tex]\[ \frac{dy}{dx} = 0 \][/tex]
Part (d): Differentiate [tex]\( x^2 + 2xy + y^2 = 8 \)[/tex] with respect to [tex]\( x \)[/tex].
Differentiate both sides with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx}(x^2 + 2xy + y^2) = \frac{d}{dx}(8) \][/tex]
The right side is straightforward:
[tex]\[ 0 \][/tex]
For the left side, differentiate each term:
[tex]\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) \][/tex]
Using product and chain rules:
[tex]\[ 2x + 2 \left( x \frac{dy}{dx} + y \right) + 2y \frac{dy}{dx} \][/tex]
Combining like terms:
[tex]\[ 2x + 2x \frac{dy}{dx} + 2y + 2y \frac{dy}{dx} \][/tex]
So, we get:
[tex]\[ 2x + 2y \frac{dy}{dx} = 0 \][/tex]
Given the context of the problem:
[tex]\[ 2x + 2y \][/tex] is the result.
In summary, the differentiated expressions are:
a) [tex]\( 2x \)[/tex]
b) [tex]\( 3 - 20x \)[/tex]
c) [tex]\( 0 \)[/tex]
d) [tex]\( 2x + 2y \)[/tex]
