# Tangent line to the curve $x^3+xy^2+x^3y^5=3$

Does the tangent line to the curve $x^3+xy^2+x^3y^5=3$ at the point $(1,1)$ pass through the point $(-2,3)$? (using implicit differentiation)

I got the implicit differentiation as $\frac{dy}{dx}=\frac{-3x^2-y^2-3x^2y^5}{2xy+5x^3y^4}$. Then I got stuck?

Explanation would be appreciated.

You've got one too many $-y^2$ terms in your numerator: you should arrive at $$\frac{dy}{dx}=\frac{-3x^2-y^2-3x^2y^5}{2xy+5x^3y^4}$$

With the correct derivative, evaluate $\dfrac {dy}{dx}$ at the point $(1, 1)$. That will give you the slope $m$ of the line tangent to your curve at $(1, 1)$. Then using $m$ and the point $(x_0, y_0) = (1, 1)$, you can write the equation of your line in point-slope form:

$$y - y_0 = m(x - x_0)\tag{1}$$

Once you've obtained your equation, then plug in the coordinates of $(-2, 3)$ into your equation of the tangent line $(1)$ to see if the equation, given $x = -2, y = 3$ is true. If so, then $(-2, 3)$ is on the line $(1)$ tangent to your curve at $(1, 1)$. If not, then it is not on that line $(1)$.

• you're right I removed the extra $y^2$ from the numerator.
– jax
Nov 19, 2013 at 18:20
• I presumed it might have simply been a "typo". Glad to help! Nov 19, 2013 at 18:23
• @amWhy: Another very busy day for you with green ruling! =` Nov 20, 2013 at 1:40
• Excellent answer as usual and it deserves $\checkmark$.
– user63181
Feb 4, 2014 at 15:55

The tangent line can be written as $y=mx+q$ where $m=\frac{\partial y}{\partial x}$ computed for $(x,y) = (1,1)$. Find $q$ by checking that the lines passes through $(1,1)$. Check if the point $(-2,3)$ satisfies the equation $y=mx+q$.