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I am working through Calculus Made Easy, by Silvanus P. Thompson, and one of the questions dealing with the chain rule threw me for a loop. I have seen recommended online to treat $y^5$ as $(y^2)^{5/2}$, but I don't quite understand how this helps, or where to go next.

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    $\begingroup$ Well, $y^5$ is not even a function of $y^2$ -- for example, when $y^2=1$, we may still have either $y^5=1$ or $y^5=-1$. So what it even means to differentiate one with respect to the other is not clear. $\endgroup$ Commented Nov 27, 2017 at 16:59
  • $\begingroup$ $y^5\neq (y^2)^{5/2}$, see the comment above. $\endgroup$
    – Levent
    Commented Nov 27, 2017 at 17:03

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You can apply the chain rule like this:

$$\frac{dy^5}{dy} = \frac{dy^5}{dy^2}\frac{dy^2}{dy}$$

$$5y^4 = \frac{dy^5}{dy^2} 2y$$

$$\frac{dy^5}{dy^2} = \frac{5y^4}{2y} = \frac{5}{2}y^3.$$

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  • $\begingroup$ I like this answer, because it helped me see better how the chain rule can be used, and how, by calculating some 'simpler' derivatives, you can manipulate the resulting equation for the answer. However, how did you decide to use $dy^5\over dy$$=$$dy^5\over dy^2$$dy^2\over dy$ in the first place? Was it by taking the desired answer $({dy^5\over dy^2})$ and manipulating it? $\endgroup$
    – oath
    Commented Nov 27, 2017 at 17:22
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    $\begingroup$ There was the chain of functions, $y\to y^2$ then $y^2\to y^5$, so it was sort of the natural thing to write down. Probably my decades of experience helped. $\endgroup$
    – B. Goddard
    Commented Nov 27, 2017 at 17:47
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To avoid a nasty definition problem, let's assume we only deal with real non-negative $y$ (the non-negative part is critical, otherwise $y^5 \ne \left(y^z\right)^{5/2}$ as has been pointed out.

Now let $z = y^2$ and note that $$ \frac{d\left[y^5\right]}{dz} = \frac{dz^{5/2}}{dz} = \frac{5}{2} z^{3/2} = \frac{5y^3}{2}. $$

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Basically you can use the property which intuitively is the chain rule ie writing a differential as the product of other differentials

$\dfrac{d (y^5)} {d(y^2) } = \dfrac{d(y^5)}{dy} * \dfrac {dy} {d(y^2)} $ which is equal to $5y^4 * \frac{1}{2y} = \frac{5}{2} y^3 $

Let me know in the comments if you wanted a solution using $(y^2)^{5/2}$

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