Question

For each problem, find the derivative of the function at the given value. Some will require a graphing calculator.

1) y= -(-x+1)^(1/2) at x= -1

2) y= 2sin (2x) at x= - π/2

3) y= x^2 + 4x + 1 at x= -1

4) y= -ln (x) at x = 2

Answer 1

1. $y= -(-x+1)^{\frac{1}{2}} \text{ at } x= -1$

$$\frac{\mathrm{d}}{\mathrm{d} x}[-\sqrt{1-x}]\\ =-\frac{\mathrm{d}}{\mathrm{d} x}[\sqrt{1-x}]\\ =-\frac{1}{2}(1-x)^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[1-x]\\ =-\frac{\frac{\mathrm{d}}{\mathrm{d} x}[1]-\frac{\mathrm{d}}{\mathrm{d} x}[x]}{2 \sqrt{1-x}}\\ =-\frac{0-1}{2 \sqrt{1-x}}\\ y'=\frac{1}{2 \sqrt{1-x}}\\[4mm] \text{ at } x=-1\\ y'=\frac{1}{2 \sqrt{2}}$$

2. $y= 2sin (2x) \text{ at } x= - π/2$

$$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{d} x}[2 \sin (2 x)] \\ =& 2 \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\sin (2 x)] \\ =& 2 \cos (2 x) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[2 x] \\ =& 2 \cos (2 x) \cdot 2 \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x] \\ =& 4 \cos (2 x) \cdot 1 \\ y'=& 4 \cos (2 x) \end{aligned} \\[4mm] \text{ at } x= -\pi/2\\ y' = 4cos(- \pi)\\ y'=4(-1)\\ y'=-4$$

3. $y= x^2 + 4x + 1 \text{ at } x= -1$

$$\frac{\mathrm{d}}{\mathrm{d} x}\left[x^{2}+4 x+1\right]\\ =\frac{\mathrm{d}}{\mathrm{d} x}\left[x^{2}\right]+4 \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[1]\\ =2 x+4 \cdot 1+0\\ y'=2 x+4 \\[4mm] \text{ at } x=-1\\ y' = 2(-1)+4\\ y' = 2$$

4. $y= -ln (x) \text{ at } x = 2$

$$\frac{\mathrm{d}}{\mathrm{d} x}[-\ln (x)]\\ =-\frac{\mathrm{d}}{\mathrm{d} x}[\ln (x)]\\ y'=-\frac{1}{x}\\[4mm] \text{at } x=2\\ y' = -\frac{1}{2}$$