Science:Math Exam Resources/Courses/MATH110/December 2013/Question 09

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MATH110 December 2013

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Question 09
Find the derivative of $y$ with respect to $x$ at the point $P=(0,\pi )$ along the curve: $\displaystyle {}\sin(x+y)=xy$

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!

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Solution

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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