Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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.

share|cite|improve this question
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 $\frac{d}{dx}[1+\tan(y)] = (1+\tan(y))'\cdot \frac{d}{dx}(y) = \sec^{2}(y)\cdot\frac{dy}{dx}$. – Alex Wertheim Aug 12 '13 at 3:35
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. – Zev Chonoles Aug 12 '13 at 3:35
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. – user89851 Aug 12 '13 at 4:32
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. – user89851 Aug 12 '13 at 4:38
up vote 1 down vote accepted

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?

share|cite|improve this answer
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. – user89851 Aug 12 '13 at 4:07
Thanks for your comment. It really helped. – user89851 Aug 12 '13 at 4:07
@user89851 Glad I could help :) – user61527 Aug 12 '13 at 4:14

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

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

share|cite|improve this answer
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. – user89851 Aug 12 '13 at 4:09

Your Answer


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.