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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.

User AndyWilson
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1 Answer

3 votes

Answer:

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)

User DJay
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7.1k points