3
$\begingroup$

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?

| cite | improve this question | |
$\endgroup$
  • 2
    $\begingroup$ The differentiation of $xy$ requires the product rule: $\frac{d}{dx}(xy)=y+xy'$. $\endgroup$ – Clayton Oct 20 '14 at 14:49
  • 1
    $\begingroup$ Consider the product rule on the term $xy$: It should read: $\frac{d}{dx}(xy) = \frac{d}{dx}x \cdot y + x \cdot \frac{d}{dx}y.$ $\endgroup$ – Jamil_V Oct 20 '14 at 14:49
  • $\begingroup$ @Clayton: AH! Fantastic find! I never saw that $\endgroup$ – Micrified Oct 20 '14 at 14:50
2
$\begingroup$

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 $y = -x + 2$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.