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:

$\begin{array}{rl}\sin (0+\pi )& =0\cdot \pi \\ \sin (\pi )& =0\\ 0& =0\end{array}$

The point is on the line. Next we want to find the derivative of the function. We apply implicit differentiation and differentiate both sides:

$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\sin (x+y)& =\frac{\mathrm{d}}{\mathrm{d}x}xy\\ \cos (x+y)(1+\frac{\mathrm{d}y}{\mathrm{d}x})& =x\frac{\mathrm{d}y}{\mathrm{d}x}+y\end{array}$

The left hand side was obtained via the chain rule and the right hand side was obtained via a product rule.

$\begin{array}{rl}\cos (x+y)+\frac{\mathrm{d}y}{\mathrm{d}x}\cos (x+y)& =x\frac{\mathrm{d}y}{\mathrm{d}x}+y\\ \frac{\mathrm{d}y}{\mathrm{d}x}\cos (x+y)-x\frac{\mathrm{d}y}{\mathrm{d}x}& =y-\cos (x+y)\\ \frac{\mathrm{d}y}{\mathrm{d}x}(\cos (x+y)-x)& =y-\cos (x+y)\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{y-\cos (x+y)}{\cos (x+y)-x}\end{array}$

Substitute in the given point.

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