question 2\nfind the equation of the line tangent to the graph of $x^{3}+y^{3}=5xy + 89$ at $(5,4)$. the…

Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $x^{3}+y^{3}=5xy + 89$ with respect to $x$. Using the power - rule and product - rule, we have: $3x^{2}+3y^{2}y'=5y + 5xy'+0$.

Step2: Solve for $y'$

Rearrange the terms to isolate $y'$: $3y^{2}y'-5xy'=5y - 3x^{2}$. Factor out $y'$: $y'(3y^{2}-5x)=5y - 3x^{2}$. So, $y'=\frac{5y - 3x^{2}}{3y^{2}-5x}$.

Step3: Find the slope of the tangent line at the point $(5,4)$

Substitute $x = 5$ and $y = 4$ into $y'$: $y'=\frac{5\times4-3\times5^{2}}{3\times4^{2}-5\times5}=\frac{20 - 75}{48 - 25}=\frac{- 55}{23}$.

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,4)$ and $m =-\frac{55}{23}$. $y - 4=-\frac{55}{23}(x - 5)$. Expand: $y-4=-\frac{55}{23}x+\frac{275}{23}$. $y=-\frac{55}{23}x+\frac{275}{23}+4$. $y=-\frac{55}{23}x+\frac{275 + 92}{23}$. $y=-\frac{55}{23}x+\frac{367}{23}$.

Answer:

$y =-\frac{55}{23}x+\frac{367}{23}$