MATH110 December 2013
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Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Take the derivative using implicit differentiation and plug in the point. Don't forget to use a product rule on the right hand side!

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution

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Before we start, let's confirm that the point $(0,\pi )$ is indeed on the line, as the statement suggests:
${\begin{aligned}\sin(0+\pi )&=0\cdot \pi \\\sin(\pi )&=0\\0&=0\end{aligned}}$
The point is on the line. Next we want to find the derivative of the function. We apply implicit differentiation and differentiate both sides:
${\begin{aligned}{\frac {\rm {d}}{{\rm {d}}x}}\sin(x+y)&={\frac {\rm {d}}{{\rm {d}}x}}xy\\\cos(x+y)\left(1+{\frac {{\rm {d}}{y}}{{\rm {d}}x}}\right)&=x{\frac {{\rm {d}}{y}}{{\rm {d}}x}}+y\end{aligned}}$
The left hand side was obtained via the chain rule and the right hand side was obtained via a product rule.
${\begin{aligned}\cos(x+y)+{\frac {{\rm {d}}{y}}{{\rm {d}}x}}\cos(x+y)&=x{\frac {{\rm {d}}{y}}{{\rm {d}}x}}+y\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}\cos(x+y)x{\frac {{\rm {d}}{y}}{{\rm {d}}x}}&=y\cos(x+y)\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}\left(\cos(x+y)x\right)&=y\cos(x+y)\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}&={\frac {y\cos(x+y)}{\cos(x+y)x}}\end{aligned}}$
Substitute in the given point.
${\begin{aligned}{\frac {{\rm {d}}{y}}{{\rm {d}}x}}(0)&={\frac {\pi \cos(0+\pi )}{\cos(0+\pi )0}}\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}(0)&={\frac {\pi (1)}{(1)0}}\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}(0)&={\frac {\pi +1}{1}}=(\pi +1)\end{aligned}}$

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MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Implicit differentiation, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

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