Given $(5x+5y~)^3= 125x^3+125y^3$, find the derivative.

Using the chain rule and power rule, I came up with

$3(5x+5y)^2 \cdot (\frac{d}{dx}5x+\frac{dy}{dx}5y)= 3 \cdot 125x^2 +3 \cdot 125y^2$

Now, the derivative of $5x~$ is 5, but what about the derivative of $5y~$?

I know that $\frac{dy}{dx}5y~$ turns to $5(\frac{dy}{dx}(y~))$

What happens after that? When I plugged the formula into Wolfram Alpha to double check my steps, it says that $\frac{dy}{dx}(y~)=0$ What is the reasoning behind that?

  • 2
    $\begingroup$ Methinks you're not understanding the operation of the derivative from this answer. When you're taking the derivative, you need to do with respect to a variable, $x$, and you need to do it to both sides of the implicit equation. On the LHS of the second line, you should have written $\frac{d}{dx} (5y)$ NOT $\frac{dy}{dx} (5y)$. And on the RHS of the same line, it should be "$3 \cdot 125 y^2 \cdot (\frac{d}{dx} (y))$. That is the chain rule. $\endgroup$ – Uticensis Mar 5 '11 at 1:26
  • $\begingroup$ The reason why we don't often write $\frac{d}{dx} (x)$, which might make the above more clear, is because it's easy to compute: the derivative of a variable with respect to itself is always 1; for every infinitesimal change in the "input", you get the exact same as the "output." $\endgroup$ – Uticensis Mar 5 '11 at 1:30
  • $\begingroup$ I think the simplest thing to do is to solve the given equation for $y$. You will get two solutions, each with a certain $x$-domain, which you can differentiate individually. You will then also have control about the $x$-domains where the resulting formulas are valid. $\endgroup$ – Christian Blatter Mar 5 '11 at 21:11

Others have pointed out that

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

because $\frac{d}{dx}(5x + 5y) = \frac{d}{dx}(5x) + \frac{d}{dx}(5y) = 5 + 5 \frac{dy}{dx}$. In your calculation when you write $\frac{dy}{dx} y$, it means $y$ multiplied with $\frac{dy}{dx}$. When you write $\frac{d}{dx}(y)$, it means $\frac{dy}{dx}$. You can think of $\frac{d}{dx}$ as a function that requires another function as input. Just like a function $f$ requires a number as input so that's why $f(x)$ doesn't mean $x$ multiplied with $f$; it means $x$ is the input to $f$.

On the right hand side, you forgot to apply the Chain Rule:

$\frac{d}{dx}(125x^3 + 125y^3) = \frac{d}{dx}(125x^3) + \frac{d}{dx}(125y^3) = 375x^2 + 375y^2 \frac{dy}{dx}$.

Finally, after this you can solve for $\frac{dy}{dx}$ as usual by isolating $\frac{dy}{dx}$ to one side.


You are applying $\frac{d}{dx}$ to both sides. On the left side, you should have $\frac{d}{dx}5y$ instead of $\frac{dy}{dx}5y$, which is just $5\frac{dy}{dx}$. And when you take $\frac{d}{dx}125y^3$ on the right, you need to use the chain rule to involve $\frac{dy}{dx}$ as well.

  • $\begingroup$ umm, you lost me on the second part. Why would I have to use the chain rule? Isn't $\frac{d}{dx}y^3= 375y^2\frac{dy}{dx}~$? $\endgroup$ – Jason Mar 5 '11 at 1:38
  • $\begingroup$ @Jason Look at your RHS; you didn't apply the differential operator to the term $125y^3$. There should be a $\frac{dy}{dx}$ term there in second line if you had. $\endgroup$ – Uticensis Mar 5 '11 at 1:40
  • $\begingroup$ @Jason: your comment above is correct, but it is not what you wrote originally. You are missing the $\frac{dy}{dx}$ factor. $\endgroup$ – Ross Millikan Mar 5 '11 at 2:50

This seems settled; I'll address how a problem like this would be handled in Mathematica (unfortunately the functionality doesn't work in Wolfram Alpha).

You'll first want to express your given in the form $f(x,y)=0$, like so:

expr = (5x + 5y)^3 - 125x^3 - 125y^3;

The key is to remember that Mathematica supports two sorts of derivatives: the partial derivative D[] and the total derivative Dt[]. As your y depends on x, Dt[] is what's appropriate here (D[] treats y as a constant, so D[y, x] gives 0).

Dt[expr, x]
-375 x^2 - 375 y^2 Dt[y, x] + 3 (5 x + 5 y)^2 (5 + 5 Dt[y, x])

From there, you can use Solve[] to solve for Dt[y, x] like so:

Dt[y, x] /. Solve[Dt[expr, x], Dt[y, x]]

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