Science:Math Exam Resources/Courses/MATH110/December 2013/Question 09/Solution 1

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.

${\begin{aligned}{\frac {{\rm {d}}{y}}{{\rm {d}}x}}(0)&={\frac {\pi -\cos(0+\pi )}{\cos(0+\pi )-0}}\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}(0)&={\frac {\pi -(-1)}{(-1)-0}}\\{\frac {{\rm {d}}{y}}{{\rm {d}}x}}(0)&={\frac {\pi +1}{-1}}=-(\pi +1)\end{aligned}}$