Question

1. find the equation of the tangent line to the curve determined by the equation

$y^{3}x+(x - y)^{4}=10x - 5y$

at the point $(1,-1)$.

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:

$$
LATEXBLOCK0
$$

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 \(m = y^\prime=\frac{7}{8}\).

$$
LATEXBLOCK1
$$

Answer:

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