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I do most of my studying independently either before I take the class to get ahead or after in order to fix trouble areas. Right now I'm trying to review Single Variable Calculus. Anyway, I ran into a road block this weekend.

Problem
Find $dy/dx$ through implicit differentiation: $$ \sin (x)=x(1+\tan(y)) $$

My solution
$$\begin{align} \frac{d}{dx}[\sin (x)]&=\frac{d}{dx}[x(1+\tan(y))]\tag{1}\\ \cos(x)&=(1)(1+ \tan(y))+x(1+\tan(y))^{-1}(\sec^{2}(y))\frac{dy}{dx}\tag{2}\\ \cos(x)&=(1+ \tan(y))+\frac{x(\sec^{2}(y))}{1+\tan(y)}\frac{dy}{dx}\tag{3}\\ \cos(x)-(1+ \tan(y))&=\frac{x(\sec^{2}(y))}{1+\tan(y)}\frac{dy}{dx}\tag{4}\\ \frac{dy}{dx}&=\frac{(\cos(x)-1- \tan(y))(1+\tan(y))}{x(\sec^{2}(y))}\tag{5} \end{align}$$

Solution from manual I'm using
$$\begin{align} \sin (x)&=x(1+\tan(y))\tag{6}\\ \cos(x) &= x(sec^{2}(y))y' + (1+\tan(y))(1)\tag{7}\\ y'&=\frac{\cos(x)-\tan(y)-1}{x\sec^{2}(y)}\tag{8} \end{align}$$

The disagreement seems to lie with steps 2 & 7. Any help figuring out why this disagreement exists would be good. Thank you for your help.

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  • $\begingroup$ Your second step is indeed incorrect. It looks like you are trying to apply an incorrect version of the power rule. Instead, apply the chain rule and note that $\frac{d}{dx}[1+\tan(y)] = (1+\tan(y))'\cdot \frac{d}{dx}(y) = \sec^{2}(y)\cdot\frac{dy}{dx}$. $\endgroup$ Aug 12, 2013 at 3:35
  • $\begingroup$ I've improved your question's formatting; apologies if I changed your meaning. You can see here how I edited your question. Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ Aug 12, 2013 at 3:35
  • $\begingroup$ AWertheim, thank you for your response. It made me take a second look at what I was doing and I believe I know what I did wrong now. I want you to know how I interpreted your comment. You wrote: $\frac{d}{dx}[1+tan(y)] = (1+\tan(y))'*\frac{d}{dx}(y)=\sec^{2}(y)$. I took this to mean: $\frac{d}{dx}[1+tan(y)] = \frac{d}{dx}(1+\tan(y))*(\frac{d}{dx}(\tan(y))+\frac{d}{dx}(1))=\sec^{2}(y)$. If that was a wrong interpretation then please elaborate. $\endgroup$
    – user89851
    Aug 12, 2013 at 4:32
  • $\begingroup$ correction AWertheim wrote: $\frac{d}{dx}[1+tan(y)] = (1+\tan(y))'*\frac{d}{dx}(y)=\sec^{2}(y)\frac{dy}{dx}$. I took this to mean: $\frac{d}{dx}[1+tan(y)] = \frac{d}{dx}(1+\tan(y))*(\frac{d}{dx}(\tan(y))+\frac{d}{dx}(1))=\sec^{2}(y) \frac{dy}{dx}$. If that was a wrong interpretation then please elaborate. $\endgroup$
    – user89851
    Aug 12, 2013 at 4:38

2 Answers 2

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Your step (2) is incorrect; the product rule states that

$$\begin{align} \frac{d}{dx} x (1 + \tan y) &= \frac{d(x)}{dx} (1 + \tan{y}) + x \frac{d(1 + \tan{y})}{dx}\\\\ &= (1 + \tan{y}) + x \left(0 + \sec^2{y} \frac{dy}{dx}\right) \end{align}$$

Where did the $(1 + \tan{y})^{-1}$ term come from?

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  • $\begingroup$ I think I realize the error from reading the comments. The extra term (1+tan(y))^-1 came from erroneously trying to resolve x*((d(1+tan(y)))/dx). I took u = 1+tan(y) and I then I differentiated as so: x* ((1+tan(y))^(1-1=-1)) * (sec(y))^2 * dy/dx instead of realizing that it should have been x* ((1+tan(y))^(1-1=0)) * (sec(y))^2 * dy/dx which resolves to x(sec(y))^2. $\endgroup$
    – user89851
    Aug 12, 2013 at 4:07
  • $\begingroup$ Thanks for your comment. It really helped. $\endgroup$
    – user89851
    Aug 12, 2013 at 4:07
  • $\begingroup$ @user89851 Glad I could help :) $\endgroup$
    – user61527
    Aug 12, 2013 at 4:14
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$$\begin{align} \frac{d}{dx}[x(1 + \tan(y))] &= (x)\left(\frac{d}{dx}[1 + \tan(y)]\right) + (1)(1 + \tan(y))\\\\ &= (x)\left(0 + \sec^2(y) \frac{dy}{dx}\right) + (1 + \tan(y))\\\\ &= x\sec^2(y) \frac{dy}{dx} + 1 + \tan(y) \end{align}$$

Essentially, you differentiated incorrectly by introducing the $(1 + \tan(y))^{-1}$ term. I'm not entirely sure where that came from.

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  • $\begingroup$ Qaphla, thank you for responding. I believe I got my answer. Your answer was similar to T.Bonger's and I only chose T.Bonger's because it was posted before. Thanks again. $\endgroup$
    – user89851
    Aug 12, 2013 at 4:09

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