# Help with Implicit Differentiation: Finding an equation for a tangent to a given point on a curve

When working through a problem set containing Implicit Differentiation problems, I've found that I keep getting the wrong answer compared to the one listed at the back of my book.

The problem is given as such: Use implicit differentiation to find the equation of the tangent line to the curve at a given point

x^2 + xy + y^2 = 3

With given point (1, 1). I also am told that it is an ellipse.

To solve this, I evidently must differentiate both sides of the problem:

1: dy/dx ( x^2 + xy + Y^2 ) = dy/dx(3)

2: dy/dx (2x + 1y'+ 2yy') = 0

3: 1y' + 2yy' = 0 - 2x

4: y'(1+2y) = -2x

5: y' = -2x/(1+2y)

Hurray, so now since I have the first derivative of Y. I can use it to find the slope at the point.

Slope at Point (1,1)= -2(1) / (1+2(1)

Slope at Point (1,1)= -2/3

So now that I've got my slope, I know the equation of the tangent will be in the form:

y=mx+b

So, given I now know the slope:

y=-2/3x + b

Substitute in the known point:

1 = -2/3(1) + b

b = 5/3

So the final answer I get is: y = -2/3x + 5/3

But according to the answer, it is supposed to be: -x + 2, I don't know where I went wrong, and I've done it twice to make sure I'm getting the same answer. Could someone please help me?

• The differentiation of $xy$ requires the product rule: $\frac{d}{dx}(xy)=y+xy'$. Oct 20, 2014 at 14:49
• Consider the product rule on the term $xy$: It should read: $\frac{d}{dx}(xy) = \frac{d}{dx}x \cdot y + x \cdot \frac{d}{dx}y.$ Oct 20, 2014 at 14:49
• @Clayton: AH! Fantastic find! I never saw that Oct 20, 2014 at 14:50

You made an error when you differentiated implicitly. You did not apply the Product Rule $(fg)' = f'g + fg'$ to the term $xy$. Keeping in mind that $y$ is a function of $x$, you should obtain

$$(xy)' = 1y + xy' = y + xy'$$

Therefore, when you differentiate

$$x^2 + xy + y^2 = 3$$

implicitly with respect to $x$, you should obtain

$$2x + y + xy' + 2yy' = 0$$

Solving for $y'$ yields

\begin{align*} xy' + 2yy' & = -2x - y\\ (x + 2y)y' & = -2x - y\\ y' & = -\frac{2x + y}{x + 2y} \end{align*}

As you can check, evaluating $y'$ at the point $(1, 1)$ yields $y' = -1$. Therefore, the tangent line equation is $y = -x + 2$.