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Problem

I have began teaching myself Calc I and I've come across the following problem:

Find $\frac{dy}{dx}$ for the following: $$x^2y+y^5\sec(x)=5.$$

I automatically presumed this was an implicit differentiation. However, since I'm somewhat new to implicit differentiation, my solution looks messy. It is below, and my question is: Am I doing this right and are there ways I could improve (whether it be in terms of my notation, method, or something else)?

Solution

\begin{align} x^2y+y^5\sec(x)&=5\\ \frac{d}{dx}\left(x^2y+y^5\sec(x)\right)&=\frac{d}{dx}5\\ \frac{d}{dx}x^2y+\frac{d}{dx}y^5\sec(x)&=\frac{d}{dx}5\\ 2xy+x^2\frac{d}{dx}(y)+\frac{d}{dx}(y^5)\sec(x)+y^5\sec(x)\tan(x)&=0\\ 2xy+x^2\frac{d}{dx}(y)+5y^4\frac{d}{dx}(y)\sec(x)+y^5\sec(x)\tan(x)&=0\\ x^2\frac{d}{dx}(y)+5y^4\frac{d}{dx}(y)\sec(x)&=-2xy-y^5\sec(x)\tan(x)\\ \frac{d}{dx}(y)\left(x^2+5y^4\sec(x)\right)&=-2xy-y^5\sec(x)\tan(x)\\ \frac{dy}{dx}&=\frac{-2xy-y^5\sec(x)\tan(x)}{x^2+5y^4\sec(x)}. \end{align}

Side question

How could I put this into Wolfram Alpha?

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  • 1
    $\begingroup$ Try "derivative of x^2*y + y^5*Sec[x] = 5" at WolframAlpha. $\endgroup$ – Amzoti Nov 21 '12 at 18:14
  • $\begingroup$ It ia fine. A bit too much detail. Two minor suggestions. Avoid $\frac{d}{dx}(y)$, use $\frac{dy}{dx}$ or $y'$. And avoid specially things like $\frac{d}{dx}(y)\sec x$, you do tat sort of thing several times. Inevitably this will be at some stage by you, or someone else, as the derivative of the product. $\endgroup$ – André Nicolas Nov 21 '12 at 18:16
  • $\begingroup$ I think I will choose $y'$ since it has the least amount of ambiguity, @AndréNicolas. $\endgroup$ – 000 Nov 21 '12 at 18:25
  • $\begingroup$ That worked perfectly. Thank you, @Amzoti. $\endgroup$ – 000 Nov 21 '12 at 18:25
  • $\begingroup$ Limitless, you are welcome. I would also recommend seeing if you can figure out how to transform the problem and arrive at some of the alternate forms that WA provided. $\endgroup$ – Amzoti Nov 21 '12 at 18:30
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$$x^2y+y^5\sec(x)=5.$$ $$2xy+x^2y'+5y^4y'\sec(x)+y^5\sec'x=0$$ $$y'(x^2+5y^4\sec x)=-2xy-y^5\frac{\sin x}{\cos^2x }$$ $$y'=\frac{dy}{dx}=\frac{-2xy-y^5\frac{\sin x}{\cos^2x }}{x^2+5y^4\sec x}$$

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  • $\begingroup$ The denominator doesn't look right. $\endgroup$ – Christopher A. Wong Nov 21 '12 at 18:37
  • $\begingroup$ I just corrected it $\endgroup$ – Adi Dani Nov 21 '12 at 18:44
  • $\begingroup$ I quite for this answer. +1 $\endgroup$ – mrs Nov 21 '12 at 19:20
  • $\begingroup$ @BabakSorouh thanks! $\endgroup$ – Adi Dani Nov 21 '12 at 19:24

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