How would I solve the following question that is troubling me.

The question is

Find an equation for the tagnet line to the graph of $x+\sin(y-2x)=1$ at point $(1,2)$

I did the following using the chain


then I simplified to $\frac{dy}{dx}=\frac{2}{cos(y-2x)}$ so I pluged in $x$ and $y $ and got $\cos(0)$ on the denominator which is $1$.

But I am unsure if I did that part correctly my final answer is $y-2=2(x-1)$

  • $\begingroup$ It is rather $1+\cos(y-2x)(\frac{dy}{dx}-2)=0$. Watch the parenthesis! $\endgroup$ – Julien Feb 21 '13 at 18:51
  • $\begingroup$ I mean I used the chain rule so its dy/dx(-2) $\endgroup$ – Fernando Martinez Feb 21 '13 at 18:58
  • $\begingroup$ I did use the chain rule too... The derivative of $y-2x$ is $(\frac{dy}{dx}-2)$. Not $\frac{dy}{dx}(-2)=-2\frac{dy}{dx}$. Can you see the difference now? $\endgroup$ – Julien Feb 21 '13 at 19:01
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    $\begingroup$ @BabakS. Hi Babak! I will wait to see if he understands my comments first. $\endgroup$ – Julien Feb 21 '13 at 19:07
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    $\begingroup$ For the chain rule, you want to multiply $\cos(y-2x)$ by the derivative of $y-2x$. This is $\frac{dy}{dx}-2$ and not $\frac{dy}{dx}(-2)$. What do you not understand? Say $y=\sin x$ for instance. Then the derivative of $y-2x=\sin x -2x$ is $\cos x-2$, and not $(\cos x)(-2)$. $\endgroup$ – Julien Feb 21 '13 at 19:13

I think what is being suggested is that:

Differentiating $$x+\sin(y-2x)=1$$ by using the chain rule should result in:

$$1 + \cos(y - 2x)\left(\frac{dy}{dx} - 2\right) = 0$$

$$\frac{dy}{dx} = -\frac{-1 + 2\cos(y - 2x)}{\cos(y - 2x)},$$ then evaluate at point $(1,2)$

  • $\begingroup$ The OP does not understand that the derivative of $y-2x$ is $\frac{dy}{dx}-2$, and not $\frac{dy}{dx}(-2)$. Maybe you can find better words than me. $\endgroup$ – Julien Feb 21 '13 at 19:15
  • $\begingroup$ I see now I was confused between subtraction sign and the negative sign which is used to multiply. $\endgroup$ – Fernando Martinez Feb 21 '13 at 19:18
  • $\begingroup$ Yes, signs can get confusing...we all slip up, with algebra, at times. Now you need only find dy/dx evaluated at $(1, 2)$, and write the tangent line as you did in your earlier problem. $\endgroup$ – Namaste Feb 21 '13 at 19:19
  • $\begingroup$ I see that distributive property was also used. $\endgroup$ – Fernando Martinez Feb 21 '13 at 19:21
  • $\begingroup$ Yes, indeed! Does the result make sense? $\endgroup$ – Namaste Feb 21 '13 at 19:23

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