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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.

5xum
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chrissy kwon
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    Use implicit differentiation. – André Nicolas Feb 18 '14 at 08:10
  • You are welcome. The implicit differentiation is mechanical, but has to be done carefully. – André Nicolas Feb 18 '14 at 08:31
  • I was wondering why the ones I've done ended up with the wrong answer, and when to know to use the implicit differentiation. – chrissy kwon Feb 18 '14 at 08:38
  • Implicit differentiations? Mistakes are easy to make, usually forgetting about the Chain Rule. – André Nicolas Feb 18 '14 at 08:39
  • how would i know, if I got the question for the first time, to use the implicit differentiation or the ones I tried? – chrissy kwon Feb 18 '14 at 08:40
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    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. – André Nicolas Feb 18 '14 at 08:47
  • Sorry for the many questions, but I thought that if you wanted a slope of a function, you need to take the derivative of it, then plug in the points, therefore ending up with a single number which would be the slope. Was I wrong? :( – chrissy kwon Feb 18 '14 at 08:53
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    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. – BlackAdder Feb 18 '14 at 08:55
  • Last question- why does just chain rule not work here when doing the steps i mentioned one above, but only implicit? – chrissy kwon Feb 18 '14 at 09:01
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    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. :) – BlackAdder Feb 18 '14 at 09:05
  • ohh no I did the same thing but the latter part was 3(xy + 1)(y+x) because derivative of xy would be (deriv of x times y) plus (x times deriv of y) and deriv of 1 (constant) would be 0 – chrissy kwon Feb 18 '14 at 09:08
  • I see, but in any case, that isn't right. Let's just focus on the derivative of $xy$ first. You have to use implicit differentiation for that. Can you differentiate $xy$ w.r.t. $x$ for me? – BlackAdder Feb 18 '14 at 09:13
  • Sorry I have no idea what w.r.t is, but I assume im supposed to write it out- i know that if I use implicit differentiation, I would get y + x dy/dx ! – chrissy kwon Feb 18 '14 at 09:15
  • That is good! So you now need to use chain rule and implicit differentiation properly. I'll edit my answer to show you the right steps. – BlackAdder Feb 18 '14 at 09:19
  • okay, I think I understand much better now though, thank you for your time and effort! It helped me alot :) – chrissy kwon Feb 18 '14 at 09:20
  • No problem! Good luck! – BlackAdder Feb 18 '14 at 09:35

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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.

BlackAdder
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  • 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. – chrissy kwon Feb 18 '14 at 08:38
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    @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! – BlackAdder Feb 18 '14 at 08:52
  • 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? – chrissy kwon Feb 18 '14 at 08:55
  • The key fact shows you HOW to perform implicit differentiation. Have you seen this done before? – BlackAdder Feb 18 '14 at 08:56
  • Yes I have, but I dont understand why theres d/dx and also dy/dx. Aren't you suppoed to use one? – chrissy kwon Feb 18 '14 at 08:57
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    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. – BlackAdder Feb 18 '14 at 08:59
  • okay, thank you for clearing it up for me :) – chrissy kwon Feb 18 '14 at 09:04
  • oops just saw this, done! :) – chrissy kwon Feb 18 '14 at 09:21