Answer:
[tex]\displaystyle \frac{dy}{dx}\Big|_{(1, 1)}=0[/tex]
Step-by-step explanation:
We are given the equation:
[tex](x^2+y^2)^2=4x^2y[/tex]
And we want to find the slope of the tangent line at the point (1,1).
Recall that the slope of the tangent line to a point of a function is given by the function's derivative.
Thus, find the derivative of the equation. Take the derivative of both sides with respect to x:
[tex]\displaystyle \frac{d}{dx}\left[(x^2+y^2)^2\right]=\frac{d}{dx}\left[4x^2y\right][/tex]
Let's do each side individually.
Left:
We can use the chain rule:
[tex](u(v(x))'=u'(v(x))\cdot v'(x)[/tex]
Let v(x) be x² + y² and u(x) is x². Thus, u'(x) is 2x. Therefore:
[tex]\displaystyle \frac{d}{dx}\left[(x^2+y^2)^2\right]=2\left(x^2+y^2\right)\left(\frac{d}{dx}\left[x^2+y^2\right]\right)[/tex]
Differentiate:
[tex]\displaystyle \frac{d}{dx}[(x^2+y^2)^2]=2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)[/tex]
Factor. Therefore, our left side is:
[tex]\displaystyle 4(x^2+y^2)\left(x+y\frac{dy}{dx}\right)[/tex]
Right:
We have:
[tex]\displaystyle \frac{d}{dx}\left[4x^2y\right][/tex]
We can move the constant multiple outside:
[tex]\displaystyle =4\frac{d}{dx}\left[x^2y\right][/tex]
We will use the Product Rule:
[tex]\displaystyle =4\left(\frac{d}{dx}\left[x^2\right]y+x^2\frac{d}{dx}\left[y\right]\right)[/tex]
Differentiate:
[tex]\displaystyle =4\left(2xy+x^2\frac{dy}{dx}\right)[/tex]
Therefore, our entire equation is:
[tex]\displaystyle 4(x^2+y^2)\left(x+y\frac{dy}{dx}\right)=4\left(2xy+x^2\frac{dy}{dx}\right)[/tex]
To find the derivative at (1,1), substitute and evaluate dy/dx when x = 1 and y = 1. Hence:
[tex]\displaystyle 4((1)^2+(1)^2)\left((1)+(1)\frac{dy}{dx}\right)=4\left(2(1)(1)+(1)^2\frac{dy}{dx}\right)[/tex]
Evaluate:
[tex]\displaystyle 4((1)+(1))\left(1+\frac{dy}{dx}\right)=4\left(2+\frac{dy}{dx}\right)[/tex]
Further simplify:
[tex]\displaystyle 8\left(1+\frac{dy}{dx}\right)=8+4\frac{dy}{dx}[/tex]
Distribute:
[tex]\displaystyle 8+8\frac{dy}{dx}=8+4\frac{dy}{dx}[/tex]
Solve for dy/dx:
[tex]\displaystyle 4\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=0[/tex]
Therefore, the slope of the tangent line at the point (1, 1) is 0.
