Question

1. find the equation of the tangent line to $x^{2}y^{3}=(x^{2}+y^{2}-1)^{3}$ at $(1,1)$.

Explanation:

Step1: Differentiate both sides implicitly

Differentiate $x^{2}y^{3}$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v = y^{3}$. So, $(x^{2}y^{3})^\prime=2xy^{3}+3x^{2}y^{2}y^\prime$.

Differentiate $(x^{2}+y^{2}-1)^{3}$ using chain - rule. Let $u=x^{2}+y^{2}-1$, then $[(x^{2}+y^{2}-1)^{3}]^\prime = 3(x^{2}+y^{2}-1)^{2}(2x + 2yy^\prime)$.

So, $2xy^{3}+3x^{2}y^{2}y^\prime=3(x^{2}+y^{2}-1)^{2}(2x + 2yy^\prime)$.

Step2: Substitute $x = 1$ and $y = 1$

Substitute $x = 1$ and $y = 1$ into the above equation:
$2\times1\times1^{3}+3\times1^{2}\times1^{2}y^\prime=3(1^{2}+1^{2}-1)^{2}(2\times1 + 2\times1\times y^\prime)$
$2 + 3y^\prime=3\times1\times(2 + 2y^\prime)$
$2+3y^\prime=6 + 6y^\prime$
$3y^\prime=-4$
$y^\prime=-\frac{4}{3}$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m =-\frac{4}{3}$.
$y - 1=-\frac{4}{3}(x - 1)$
$3y-3=-4x + 4$
$4x+3y=7$.

Answer:

$4x + 3y=7$