2. find the equation of the tangent line to the curve determined by the equation\n\n$y^{3}x+(x - y)^{4}=10x…

Explanation:

Step1: Differentiate both sides

Differentiate (y^{3}x+(x - y)^{4}=10x-5y) with respect to (x) using product rule ((uv)^\prime = u^\prime v+uv^\prime) and chain rule ((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)).

For the left - hand side:

• The derivative of (y^{3}x) is (y^{3}+3xy^{2}y^\prime) (by product rule, where (u = x), (v = y^{3}), (u^\prime=1), (v^\prime = 3y^{2}y^\prime)).
• The derivative of ((x - y)^{4}) is (4(x - y)^{3}(1 - y^\prime)) (by chain rule, let (u=x - y), (f(u)=u^{4}), (f^\prime(u) = 4u^{3}), (u^\prime=1 - y^\prime)).

For the right - hand side:

• The derivative of (10x-5y) is (10-5y^\prime).

So, (y^{3}+3xy^{2}y^\prime+4(x - y)^{3}(1 - y^\prime)=10-5y^\prime).

Step2: Substitute (x = 1) and (y=-1)

Substitute (x = 1) and (y=-1) into the above equation:

[ \begin{align*} (-1)^{3}+3\times1\times(-1)^{2}y^\prime+4(1-(-1))^{3}(1 - y^\prime)&=10-5y^\prime\ -1 + 3y^\prime+4\times8(1 - y^\prime)&=10-5y^\prime\ -1+3y^\prime + 32-32y^\prime&=10-5y^\prime\ 31-29y^\prime&=10-5y^\prime\ -29y^\prime + 5y^\prime&=10 - 31\ -24y^\prime&=-21\ y^\prime&=\frac{7}{8} \end{align*} ]

Step3: Use the point - slope form

The point - slope form of a line is (y - y_{0}=m(x - x_{0})), where ((x_{0},y_{0})=(1,-1)) and (m = y^\prime=\frac{7}{8}).

[ \begin{align*} y-(-1)&=\frac{7}{8}(x - 1)\ y + 1&=\frac{7}{8}x-\frac{7}{8}\ y&=\frac{7}{8}x-\frac{7}{8}-1\ y&=\frac{7}{8}x-\frac{15}{8} \end{align*} ]

Answer:

(y=\frac{7}{8}x-\frac{15}{8})