(X^3)y + (x^2)y = 5xy find dy/dx @(2,5) and the equation of the tangent line.
You have to use implicit differentiation.
$\displaystyle yx^3+yx^2=5xy$
Apply the product rule to each term. Since each term is a product of two functions, the derivative will have 6 terms:
$\displaystyle (x^3\frac{dy}{dx}+3x^2y)+(\frac{dy}{dx}x^2+2xy)=5x \frac{dy}{dx}+5y$
Now just do the algebra to solve for $\displaystyle \frac{dy}{dx}$. Then evaluate the derivative at that point. Just plug in the numbers (2,5) in the expression for the derivative. That will be the slope of your tangent line.
The derivative I came up with looks like this:
$\displaystyle \frac{dy}{dx}=\frac{5y-3x^2y+2xy}{x^3+x^2-5x}$
Make sure you can get the same thing. Just plug in the x's and the y's. Then use the fact that the point (2,5) is on the line to wright the equation from $\displaystyle y=mx+b$