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I am a bit confused about taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$.

$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2 y^2)}{dx} $$

Edit: Incorrect step $$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 10xy^2 + 5x^22y\frac{dy}{dx}$$ $$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 5 \left(2xy^2 + x^22y\frac{dy}{dx}\right)$$ ...

So I don't get how we got $2y\frac{dy}{dx}$ from $y^2$. Why it is not $2\frac{dy}{dx}$???

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  • $\begingroup$ Are the *s meant to be multiplication? In addition, you differentiate both sides with respect to $x$, not multiply it by $\frac{dy}{dx}$. I will show you the correction in my answer. $\endgroup$
    – Shuri2060
    Feb 25, 2016 at 18:04
  • $\begingroup$ @QuestionAsker yes * is multiplication $\endgroup$
    – YohanRoth
    Feb 25, 2016 at 18:05
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    $\begingroup$ DO NOT write $\displaystyle \frac{dy}{dx} [(x^2+y^2)^3]$ when you mean $\displaystyle \frac{d}{dx} [(x^2+y^2)^3]$. Those are two entirely different things. $\qquad$ $\endgroup$ Feb 25, 2016 at 18:06
  • $\begingroup$ @MichaelHardy I just copied straight from Khan Academy... Sal writes as is $\endgroup$
    – YohanRoth
    Feb 26, 2016 at 0:36
  • $\begingroup$ MaharahaX : Do you have a link? $\qquad$ $\endgroup$ Feb 26, 2016 at 4:35

4 Answers 4

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Well, basically it is simple chain rule. So, what is $(y^{2})'$? It is simply $$(f(x)^{2})'$$ which looks like simple chain rule. Just derivate the entire function to get $2f(x)$ and multiply with the inner function's derivative which is $f'(x)$. So you get $$2f(x)f'(x)$$ Which is simply $$2y\frac{dy}{dx}$$

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  • $\begingroup$ is here $y=y(x)$? $\endgroup$ Feb 25, 2016 at 18:25
  • $\begingroup$ Yes, $y$ is assumed to be dependent upon $x$ only $\endgroup$
    – Socre
    Feb 25, 2016 at 18:27
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    $\begingroup$ When I was teaching, I'd have students replace every occurence of $y$ by $f(x)$ and once the derivatives taken to come back to $y$. They hated doing this, but those who accepted to work this way performed much better than those who didn't. $\endgroup$
    – MasB
    Feb 25, 2016 at 18:58
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The 2nd line should be:

$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2y^2)}{dx}$$

I think this is the key to why you are not understanding how to implicitly differentiate that term.

When you implicitly differentiate, you differentiate both sides of the equation with respect to $x$ (or whatever variable suits you). This is "doing" $\frac{d}{dx}$ to both sides.

To implicitly differentiate $x=y^2$, for example, you do this:

Let: $x=y^2$

$$\frac{d(x)}{dx}=\frac{d(y^2)}{dx}$$

$$1=1\times\frac{d(y^2)}{dx}$$

$$1=\frac{dy}{dy}\times\frac{d(y^2)}{dx}$$

Here, you are allowed to swap the bottoms of the fractions, just like if you had $\frac{1}{3}\times\frac{4}{2}$.

$$1=\frac{dy}{dx}\times\frac{d(y^2)}{dy}$$

Now you can differentiate $y^2$ like you "normally" do as you are differentiating with respect to the same variable, $y$.

$$1=2y\frac{dy}{dx}$$

The key to implicit differentiation is that you can only differentiate (as in doing what you normally do - ie. $x^2$ goes to $2x$ when differentiated) a function containing a variable with respect to that variable. To solve the problem of differentiating a function with $y$s in with respect to $x$, it may help to multiply the fraction by "1" (ie. $\frac{dy}{dy}$) as shown above.

So: $$\frac{d(x^2)}{dx}=2x$$

But:

$$\frac{d(y^2)}{dx} ≠ 2x$$

Normally, when you differentiate a function like $y=x^3$, you're still "doing the same thing to both sides":

$$y=x^3$$

$$\frac{d(y)}{dx}=\frac{d(x^3)}{dx}$$

$$\frac{dy}{dx}=3x^2$$

If the thing you are trying to differentiate has more than one kind of variable in it, like $xy$, then you can still use the normal differentiation rules:

Eg. Let $w=xy$. Find $\frac{dw}{dx}$.

$$w=xy$$

$$\frac{dw}{dx}=\frac{d(xy)}{dx}$$

You can use the product rule in the same way as before by letting $u = x$ and $v = y$.

Normally, $z=uv \implies z'=u'v+uv'$.

So: $$\frac{dw}{dx}=\frac{d(xy)}{dx}=x'y + xy'$$

$$\frac{dw}{dx}=\frac{dx}{dx}y + x\frac{dy}{dx}$$

$$\frac{dw}{dx}=y + x\frac{dy}{dx}$$

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  • $\begingroup$ I changed my example to make it clear that when you implicitly differentiate, you are still doing the same thing to both sides of the equation, or multiplying one side by 1. $\endgroup$
    – Shuri2060
    Feb 25, 2016 at 18:23
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I hate the way implicit derivatives are taught, but rather than boring you with my own method, the short answer is this - The derivative of $x^2$ is $2x\frac{dx}{dx}$. For the same reason, the derivative of $y^2$ is $2y\frac{dy}{dx}$. It's just that on the first one, since $\frac{dx}{dx}$ is 1, it goes away, leaving you with only $2x$.

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  • $\begingroup$ Yes - I made the same confusion when I was first taught implicit. I think the key to avoiding it is learning that with differentiation, you are still "doing the same thing to both sides". $\endgroup$
    – Shuri2060
    Feb 25, 2016 at 18:34
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An expression $f(x,y)=g(x,y)$ is implicitly differentiated by

$$ \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial x} x' =\frac{\partial g}{\partial y} y' + \frac{\partial g}{\partial x} x'$$

but with $x'=1$ in your case.

What I get is

$$ \left(6 y (x^2+y^2)^2 \right) y' + \left( 6 x (x^2 + y^2)^2 \right) 1 = \left( 10 x^2 y \right) y' + \left(10 x y^2 \right) 1$$

which needs some simplifications.

The specific question as to why $\frac{{\rm d} }{{\rm d} x}y^2 = 2 y y'$ use the rule above to get

$$ \frac{\partial y^2}{\partial y} y' + \frac{\partial y^2}{\partial x} 1 = (2 y) y' + (0) 1 = 2 y y' $$

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