Implicit differentiation of $\sin(x)=x(1+\tan(y))$

Question

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.

Answer 1

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?

Answer 2

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

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