Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Try "derivative of x^2*y + y^5*Sec[x] = 5" at WolframAlpha. –  Amzoti Nov 21 '12 at 18:14
    
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. –  André Nicolas Nov 21 '12 at 18:16
    
I think I will choose $y'$ since it has the least amount of ambiguity, @AndréNicolas. –  000 Nov 21 '12 at 18:25
    
That worked perfectly. Thank you, @Amzoti. –  000 Nov 21 '12 at 18:25
    
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. –  Amzoti Nov 21 '12 at 18:30
add comment

1 Answer 1

up vote 2 down vote accepted

$$x^2y+y^5\sec(x)=5.$$ $$2xy+x^2y'+5y^4y'\sec(x)+y^5\sec'x=0$$ $$y'(x^2+5y^4\sec x)=-2xy-y^5\frac{\sin x}{\cos^2x }$$ $$y'=\frac{dy}{dx}=\frac{-2xy-y^5\frac{\sin x}{\cos^2x }}{x^2+5y^4\sec x}$$

share|improve this answer
    
The denominator doesn't look right. –  Christopher A. Wong Nov 21 '12 at 18:37
    
I just corrected it –  Adi Dani Nov 21 '12 at 18:44
    
I quite for this answer. +1 –  B. S. Nov 21 '12 at 19:20
    
@BabakSorouh thanks! –  Adi Dani Nov 21 '12 at 19:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.