Question

Use implicit differentiation to find the slope of the tangent line at the given point:

Answer 1

Hey there!

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Given:

Bifolium (x²+y²)² = 4x²y

Point (1, 1)

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Solution:

4x³ + 4x²y * 1 * dy/dx + 4xy² + 4y³ * 1 * dy/dx = 8xy + 4x² * dy/dx

~Calculate the products

4x³ + 4x²y * dy/dx + 4xy² + 4y³ * dy/dx = 8xy + 4x² * dy/dx

~Move terms

4x²y * dy/dx + 4y³ * dy/dx - 4x² * dy/dx = 8xy - 4x³ - 4xy²

~Factor

(4x²y + 4y³ - 4x²) * dy/dx = 8xy - 4x³ - 4xy²

~Divide both sides

dy/dx = 8xy-4x³-4xy²/4x²y+4y³-4x²

~Simplify

dy/dx = 2xy-x³-xy²/x²y+y³-x²

~Solve using the point given

dy/dx = 2(1)(1)-(1)³-(1)(1)²/(1)²(1)+(1)³-(1)²

dy/dx = 2*1-1-1/1*1+1-1

dy/dx = 0/1

dy/dx = 0

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Answer:

dy/dx = 0

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Best of Luck!

Answer 2

Answer:

$\frac{dy}{dx}=0$

Step-by-step explanation:

So we have the equation:

$(x^2+y^2)^2=4x^2y$

And we want to find the slope of the tangent line at the point (1,1).

So, let's implicitly differentiate. Take the derivative of both sides:

$\frac{d}{dx}[(x^2+y^2)^2]=\frac{d}{dx}[4x^2y]$

Let's do each side individually.

Left:

We can use the chain rule:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let's let v(x) be x²+y². So, u(x) is x². Thus, the u'(x) is 2x. Therefore:

$\frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)(\frac{d}{dx}[x^2+y^2])$

We can differentiate x like normal. However, for y, we must differentiate implicitly. pretend y is y(x). This gives us:

$\frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)(\frac{d}{dx}[x^2]+\frac{d}{dx}[y^2(x)])$

Differentiate:

$\frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)(2x+2y\frac{dy}{dx})$

Therefore, our left side is:

$2(x^2+y^2)(2x+2y\frac{dy}{dx})$

Right:

We have:

$\frac{d}{dx}[4x^2y]$

Let's move the 4 outside:

$=4\frac{d}{dx}[x^2y]$

Use the product rule:

$=4(\frac{d}{dx}[x^2]y+x^2\frac{d}{dx}[y])$

Differentiate:

$=4(2xy+x^2\frac{dy}{dx})$

Therefore, our entire equation is:

$2(x^2+y^2)(2x+2y\frac{dy}{dx})=4(2xy+x^2\frac{dy}{dx})$

So, to find the derivative at (1,1), substitute 1 for x and 1 for y.

$2((1)^2+(1)^2)(2(1)+2(1)\frac{dy}{dx})=4(2(1)(1)+(1)^2\frac{dy}{dx})$

Evaluate.

$2((1)+(1))(2+2\frac{dy}{dx})=4(2+\frac{dy}{dx})$

Simplify. Also, let's distribute the right:

$2(2)(2+2\frac{dy}{dx})=8+4\frac{dy}{dx}$

Multiply.

$4(2+2\frac{dy}{dx})=8+4\frac{dy}{dx}$

Distribute the left:

$8+8\frac{dy}{dx}=8+4\frac{dy}{dx}$

Subtract 8 from both sides:

$8\frac{dy}{dx}=4\frac{dy}{dx}$

Subtract 4(dy/dx) from both sides:

$4\frac{dy}{dx}=0$

Divide both sides by 4:

$\frac{dy}{dx}=0$

Therefore, the slope at the point (1,1) is 0.

And we're done!

We can verify this using the graph. The slope of the line tangent to the point (1,1) seems like it would be horizontal, giving us a slope of 0.

Edit: Typo