Answer:
Part A)
[tex]\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}[/tex]
Part B)
[tex]\displaystyle y=-\frac{5}{8}x+\frac{9}{4}[/tex]
Step-by-step explanation:
We have the equation:
[tex]\displaystyle x^2y+y^2x=6[/tex]
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:
[tex]\displaystyle \frac{d}{dx}\Big[x^2y+y^2x\Big]=\frac{d}{dx}\big[6\big][/tex]
The derivative of a constant is 0. We can expand the left:
[tex]\displaystyle \frac{d}{dx}\Big[x^2y\Big]+\frac{d}{dx}\Big[y^2x\Big]=0[/tex]
Differentiate using the product rule:
[tex]\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[/tex]
Implicitly differentiate:
[tex]\displaystyle (2xy+x^2\frac{dy}{dx})+(2y\frac{dy}{dx}x+y^2)=0[/tex]
Rearrange:
[tex]\displaystyle \Big(x^2\frac{dy}{dx}+2xy\frac{dy}{dx}\Big)+(2xy+y^2)=0[/tex]
Isolate the dy/dx:
[tex]\displaystyle \frac{dy}{dx}(x^2+2xy)=-(2xy+y^2)[/tex]
Hence, our derivative is:
[tex]\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}[/tex]
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:
[tex]\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{2(2)(1)+(1)^2}{(2)^2+2(2)(1)}[/tex]
Evaluate:
[tex]\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{4+1}{4+4}=-\frac{5}{8}[/tex]
Then by the point-slope form:
[tex]y-y_1=m(x-x_1)[/tex]
Yields:
[tex]\displaystyle y-1=-\frac{5}{8}(x-2)[/tex]
Distribute:
[tex]\displaystyle y-1=-\frac{5}{8}x+\frac{5}{4}[/tex]
Hence, our equation is:
[tex]\displaystyle y=-\frac{5}{8}x+\frac{9}{4}[/tex]
