Question

A) Findi

Answer 1

Answer:

Part A)

$\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}$

Part B)

$\displaystyle y=-\frac{5}{8}x+\frac{9}{4}$

Step-by-step 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:

$\displaystyle \frac{d}{dx}\Big[x^2y+y^2x\Big]=\frac{d}{dx}\big[6\big]$

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

$\displaystyle \frac{d}{dx}\Big[x^2y\Big]+\frac{d}{dx}\Big[y^2x\Big]=0$

Differentiate using the product rule:

$\displaystyle \Big(\frac{d}{dx}\big[x^2\big]y+x^2\frac{d}{dx}\big[y\big]\Big)+\Big(\frac{d}{dx}\big[y^2\big]x+y^2\frac{d}{dx}\big[x\big]\Big)=0$

Implicitly differentiate:

$\displaystyle (2xy+x^2\frac{dy}{dx})+(2y\frac{dy}{dx}x+y^2)=0$

Rearrange:

$\displaystyle \Big(x^2\frac{dy}{dx}+2xy\frac{dy}{dx}\Big)+(2xy+y^2)=0$

Isolate the dy/dx:

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

Hence, our derivative is:

$\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}$

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:

$\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{2(2)(1)+(1)^2}{(2)^2+2(2)(1)}$

Evaluate:

$\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{4+1}{4+4}=-\frac{5}{8}$

Then by the point-slope form:

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

Yields:

$\displaystyle y-1=-\frac{5}{8}(x-2)$

Distribute:

$\displaystyle y-1=-\frac{5}{8}x+\frac{5}{4}$

Hence, our equation is:

$\displaystyle y=-\frac{5}{8}x+\frac{9}{4}$