1
$\begingroup$

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

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

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 $y-2=2(x-1)$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @BabakS. Hi Babak! I will wait to see if he understands my comments first. $\endgroup$ – Julien Feb 21 '13 at 19:07
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $(1,2)$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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