0
$\begingroup$

enter image description here

I hope the picture's quality isn't too bad, but the question was slope of line tangent to $y^2 + (xy +1) ^3 = 0$ at point $(2,-1)$.

I tried two ways where one was trying to find the derivative right away- is this way wrong? especially the second part where I put $2(xy + 1) ^2$ ?

The second way I tried was actually multiplying it out because I was sort of hopeless. Bot hare wrong answers, and the slope was $\frac34$.

Please help me.

$\endgroup$
16
  • 3
    $\begingroup$ Use implicit differentiation. $\endgroup$ Feb 18, 2014 at 8:10
  • $\begingroup$ You are welcome. The implicit differentiation is mechanical, but has to be done carefully. $\endgroup$ Feb 18, 2014 at 8:31
  • 1
    $\begingroup$ You cannot pleasantly solve for $y$ in terms of $x$. Being able to solve for $x$ in terms of $y$ would also be good, you could calculate $\frac{dx}{dy}$ and take the reciprocal. But you (and I) can't do that either. So implicit is the only possibility. Actually, I use implicit even when I can solve. It is easier to go from $x^2+y^2=25$ to $2x+2y\frac{dy}{dx}=0$ than to solve for $y$ and differentiate. $\endgroup$ Feb 18, 2014 at 8:47
  • 1
    $\begingroup$ This is correct. But finding the derivative $\dfrac{dy}{dx}$ is not as straight-forward as you think. As I mentioned in my comment, you seem to be trying to hold $y$ constant, which is a very common mistake to make. $\endgroup$
    – BlackAdder
    Feb 18, 2014 at 8:55
  • 1
    $\begingroup$ I'm assuming you want to perform the chain rule on $(xy+1)^3$? If so, then if you were to do $\frac{d}{dx}(xy+1)^3=3(xy+1)^2(y+1)$, then can you see that you are treating $y$ as a constant? Then, from the previous comments, there is your mistake. :) $\endgroup$
    – BlackAdder
    Feb 18, 2014 at 9:05

1 Answer 1

1
$\begingroup$

Function: $y^2 + (xy +1) ^3 = 0$

Differentiate implicitly: \begin{align*} y^2 + (xy +1) ^3 &= 0\\ \implies \frac{d}{dx}y^2 + (xy +1) ^3 &= 0\\ \implies 2y\frac{dy}{dx}+3(xy+1)^2\left(y+x\frac{dy}{dx}\right)&=0 \end{align*}

Now substitute $(x,y)$ and $\dfrac{dy}{dx}$ the subject.

Alternatively, you could make $y$ the subject in your original function.

Key Fact:

$$\frac{d}{dx}y^2=\frac{dy}{dx}\frac{d}{dy}y^2=\frac{dy}{dx}2y$$

Edit (Chain rule and implicit differentiation): \begin{align*} \frac{d}{dx} (xy +1) ^3 &= 0\\ 3(xy+1)^2\times\frac{d}{dx}(xy+1)&= 0 \end{align*} Note that for the chain rule, you just bring down the power from outside the bracket and reduce the power by one then perform the differentiation on the inside.

$\endgroup$
8
  • $\begingroup$ I was wondering why the ones I've done ended up with the wrong answer, and also how I know why I need to use implicit differentiation. $\endgroup$ Feb 18, 2014 at 8:38
  • 1
    $\begingroup$ @chrissykwon Unfortunately, I can't see your working. :( But the key observation is that you cannot treat $y$ as a constant because it is always varying with $x$. Hence, implicit differentiation is the way to go. Hope that answered your question. If you have another question, go for it! $\endgroup$
    – BlackAdder
    Feb 18, 2014 at 8:52
  • $\begingroup$ Thank you for the first answer! I think I get why now, but I don't understand what the Key Fact you wrote means. Can you explain it a bit? $\endgroup$ Feb 18, 2014 at 8:55
  • $\begingroup$ The key fact shows you HOW to perform implicit differentiation. Have you seen this done before? $\endgroup$
    – BlackAdder
    Feb 18, 2014 at 8:56
  • 1
    $\begingroup$ Hmm, this is a little tricky here, but the $\frac{d}{dx}$ here is meant to be an operator, that means it does something to the LHS or RHS, while $\frac{dy}{dx}$ is the variable. Perhaps your teacher uses a different notation, in that case, you should probably ignore what I've written about the key fact. $\endgroup$
    – BlackAdder
    Feb 18, 2014 at 8:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .