The slope of the tangent to the curve $10x^3+5x^2y+4xy^2+6y^3=25$

Question

I am having problems understanding how to solve this equation:

The slope of the tangent to the curve: $$10x^3+5x^2y+4xy^2+6y^3=25$$ at the point $(1,1)$.

Any help would be appreciated, thanks!

Answer 1

Hint: $$\dfrac{dy}{dx}=-\dfrac{F'_{x}}{F'_{y}}=\dfrac{30x^2+10xy+4y^2}{5x^2+8xy+18y^2}$$ so $$y'_{x}|_{x=1}=-\dfrac{44}{31}$$ then the tangent $$y-1=-\dfrac{44}{31}(x-1)$$ where $$F=10x^3+5x^2y+4xy^2+6y^3-25$$

Answer 2

After differentiating and finding an expression for $y'$, once you put the given values of $(1,1)$ for $(x,y)$, your answer will be $-44/31$.

Answer 3

Differentiating wrt $x$, $$30x^2+\left(10xy+5x^2\frac{dy}{dx}\right)+\left(4y^2+8xy\frac{dy}{dx}\right)+18y^2\dfrac{dy}{dx}=0$$ At $(1, 1)$, this gives $30+10+5y'+4+8y'+18y'=0$, so what is $y'$?