If 5x^2+5x+xy=2 and y(2)= -14 find y'(2) by implicit differentiation
5 x² + 5 x + x y = 2
Differentiate both sides of the equation with respect to x.
Apply the sum / difference rule: ( f ± g )′ = f ′ ± g′
( 5 x² )′ + ( 5 x )′ + ( x y )′ = 2′
5 ∙ 2 x + 5 + ( x y )′ = 0
10 x + 5 + ( x y )′ = 0
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For ( x y )′ apply the product rule:
( f ∙ g )′ = f ′ ∙ g + f ∙ g′
( x y )′ = x ′ ∙ y + x ∙ y′ (x)
( x y )′ = 1 ∙ y + x ∙ y′ (x)
( x y )′ = y + x ∙ y′ (x)
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10 x + 5 + y + x ∙ y′ (x) = 0
y(2) = -14 mean x = 2 , y = - 14
10 ∙ 2 + 5 + ( - 14 ) + 2 ∙ y′ ( 2 ) = 0
20 + 5 - 14 + 2 ∙ y′ ( 2 ) = 0
11 + 2 ∙ y′ ( 2 ) = 0
2 ∙ y′ ( 2 ) = - 11
y′ ( 2 ) = - 11 / 2
Well, well, well, we meet again with these implicit differentiation shenanigans! Let's dive right into it, shall we?
Since we want to find y'(2), first, let's implicitly differentiate the equation 5x^2 + 5x + xy = 2 with respect to x. Hold on tight, things are about to get ridiculous!
Applying the product rule to the xy term, we get:
5x^2 + 5x + xy = 2
Differentiating both sides with respect to x gives us:
(10x + 5) + (x * dy/dx + y) = 0
Now, we can try to isolate dy/dx on one side, because that's what we're after. Alright, spruce up your math hats and let's do some rearranging:
x * dy/dx + (10x + y) + 5 = 0
And finally, solving for dy/dx, we are left with:
dy/dx = -(10x + y + 5) / x
Now, to find y'(2), we substitute x = 2 into our equation:
dy/dx = -(10*2 + y + 5) / 2
Plugging in the value y(2) = -14:
dy/dx = -(20 + (-14) + 5) / 2
dy/dx = -11 / 2
And there you have it, my friend! y'(2) equals -11/2. Don't let this crazy math circus scare you off!
To find y'(2) using implicit differentiation, we'll differentiate both sides of the equation 5x^2 + 5x + xy = 2 with respect to x.
For the term 5x^2, we apply the power rule which states that the derivative of x^n is nx^(n-1). In this case, n = 2:
d/dx (5x^2) = 10x.
For the term 5x, which is equivalent to 5x^1, we again apply the power rule:
d/dx (5x) = 5.
For the term xy, we need to use the product rule. The product rule states that if u and v are functions of x, then the derivative of their product is given by:
d/dx (u * v) = u' * v + u * v'.
Using the product rule, we have:
d/dx (xy) = x * (d/dx (y)) + (d/dx (x)) * y.
Since we are differentiating with respect to x, y is considered a function of x. Therefore, d/dx (y) = y'.
Putting it all together, we have:
10x + 5 + x * y' + 5y = 0.
Now, we substitute x = 2 and y(2) = -14 into the equation:
10(2) + 5 + 2 * y' + 5(-14) = 0.
Simplifying, we get:
20 + 5 - 140 + 2y' = 0.
-115 + 2y' = 0.
2y' = 115.
y' = 115/2.
Therefore, y'(2) = 115/2 (or 57.5).
To find y'(2) by implicit differentiation, we need to differentiate both sides of the equation with respect to x and solve for dy/dx, which represents the derivative of y with respect to x.
Let's start with the given equation:
5x^2 + 5x + xy = 2
Now, we differentiate both sides of the equation with respect to x term by term. Remember to use the product rule when differentiating terms that involve y:
d/dx(5x^2 + 5x + xy) = d/dx(2)
Differentiating the left side:
d/dx(5x^2) + d/dx(5x) + d/dx(xy) = 0
Simplifying:
10x + 5 + x(dy/dx) + y = 0
Next, let's rearrange the equation to isolate dy/dx:
x(dy/dx) + y = -10x - 5
Now, separate the terms involving dy/dx:
x(dy/dx) = -10x - 5 - y
Next, divide both sides by x:
dy/dx = (-10x - 5 - y) / x
Now, we need to find y when x = 2 in order to calculate y'(2). We are given that y(2) = -14, so substitute these values into the equation:
y'(2) = (-10(2) - 5 - (-14)) / 2
Simplify further:
y'(2) = (-20 - 5 + 14) / 2
y'(2) = -11 / 2
Therefore, y'(2) = -11/2.