Question

A) Findi

Consider the curve x²y + y2x = 6
dy
in terms of x and y
B) Write the equation for the tangent line where x = 2 and y = 1.

Answer 1

Part A)

`\displaystyle (dy)/(dx)=-(2xy+y^2)/(x^2+2xy)`

Part B)

`\displaystyle y=-(5)/(8)x+(9)/(4)`

Explanation:

We have the equation:

`\displaystyle x^2y+y^2x=6`

Part A)

We want to find the derivative of our function, dy/dx.

So, we will take the derivative of both sides with respect to x:

`\displaystyle (d)/(dx)\Big[x^2y+y^2x\Big]=(d)/(dx)\big[6\big]`

The derivative of a constant is 0. We can expand the left:

`\displaystyle (d)/(dx)\Big[x^2y\Big]+(d)/(dx)\Big[y^2x\Big]=0`

Differentiate using the product rule:

`\displaystyle \Big((d)/(dx)\big[x^2\big]y+x^2(d)/(dx)\big[y\big]\Big)+\Big((d)/(dx)\big[y^2\big]x+y^2(d)/(dx)\big[x\big]\Big)=0`

Implicitly differentiate:

`\displaystyle (2xy+x^2(dy)/(dx))+(2y(dy)/(dx)x+y^2)=0`

Rearrange:

`\displaystyle \Big(x^2(dy)/(dx)+2xy(dy)/(dx)\Big)+(2xy+y^2)=0`

Isolate the dy/dx:

`\displaystyle (dy)/(dx)(x^2+2xy)=-(2xy+y^2)`

Hence, our derivative is:

`\displaystyle (dy)/(dx)=-(2xy+y^2)/(x^2+2xy)`

Part B)

We want to find the equation of the tangent line at (2, 1).

So, let's find the slope of the tangent line using the derivative. Substitute:

`\displaystyle (dy)/(dx)_((2,1))=-(2(2)(1)+(1)^2)/((2)^2+2(2)(1))`

Evaluate:

`\displaystyle (dy)/(dx)_((2,1))=-(4+1)/(4+4)=-(5)/(8)`

Then by the point-slope form:

`y-y_1=m(x-x_1)`

Yields:

`\displaystyle y-1=-(5)/(8)(x-2)`

Distribute:

`\displaystyle y-1=-(5)/(8)x+(5)/(4)`

Hence, our equation is:

`\displaystyle y=-(5)/(8)x+(9)/(4)`