Question

1. $x^{3}y+xy^{3}=-8$

for the following exercises, find the equation of the tangent line to the graph of the given equation at the indicated
point. use a calculator or computer software to graph the function and the tangent line.

1. t $x^{4}y-xy^{3}=-2,(-1,-1)$

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{4}\) and \(v = y\) for the term \(x^{4}y\) (so \((x^{4}y)^\prime=4x^{3}y + x^{4}y^\prime\)), and for the term \(xy^{3}\) with \(u = x\) and \(v = y^{3}\) (so \((xy^{3})^\prime=y^{3}+3xy^{2}y^\prime\)).
Differentiating \(x^{4}y-xy^{3}=-2\) gives:
\(4x^{3}y + x^{4}y^\prime-(y^{3}+ 3xy^{2}y^\prime)=0\)

Step2: Solve for \(y^\prime\)

Expand the equation: \(4x^{3}y + x^{4}y^\prime - y^{3}-3xy^{2}y^\prime = 0\)
Group the terms with \(y^\prime\): \(y^\prime(x^{4}-3xy^{2})=y^{3}-4x^{3}y\)
Then \(y^\prime=\frac{y^{3}-4x^{3}y}{x^{4}-3xy^{2}}\)

Step3: Substitute \(x = - 1,y=-1\) into \(y^\prime\)

\(y^\prime=\frac{(-1)^{3}-4(-1)^{3}(-1)}{(-1)^{4}-3(-1)(-1)^{2}}\)
First, calculate the numerator: \((-1)^{3}-4(-1)^{3}(-1)=-1 - 4=-5\)
Second, calculate the denominator: \((-1)^{4}-3(-1)(-1)^{2}=1 + 3=4\)
So \(y^\prime=-\frac{5}{4}\)

Step4: Use the point - slope form \(y - y_{0}=m(x - x_{0})\)

Here \(x_{0}=-1,y_{0}=-1,m =-\frac{5}{4}\)
\(y+1=-\frac{5}{4}(x + 1)\)
Expand it: \(y=-\frac{5}{4}x-\frac{5}{4}-1=-\frac{5}{4}x-\frac{9}{4}\)

Answer:

\(y =-\frac{5}{4}x-\frac{9}{4}\)