(6 points) Use the definition of derivatives to prove that if: 1. f(x) = x2 + 6x − 5 then f'(x) = 2x + 6. ii. g(x) = csc x then g'(x) = -cot x csc x.

$$f(x+h) = (x+h)^2 + 6(x+h) - 5$$ $$f(x+h) = x^2 + 2xh + h^2 + 6x + 6h - 5$$

$$f(x+h) - f(x) = (x^2 + 2xh + h^2 + 6x + 6h - 5) - (x^2 + 6x - 5)$$ $$f(x+h) - f(x) = x^2 + 2xh + h^2 + 6x + 6h - 5 - x^2 - 6x + 5$$ $$f(x+h) - f(x) = 2xh + h^2 + 6h$$

$$f'(x) = \lim_{h \to 0} \frac{2xh + h^2 + 6h}{h}$$

Factor out $h$ from the numerator:

$$f'(x) = \lim_{h \to 0} \frac{h(2x + h + 6)}{h}$$

Cancel $h$ (since $h \neq 0$ in the limit process):

$$f'(x) = \lim_{h \to 0} (2x + h + 6)$$

$$f'(x) = 2x + 0 + 6$$ $$\boxed{f'(x) = 2x + 6}$$

$$g(x+h) - g(x) = \csc(x+h) - \csc(x)$$

Convert to sines:

$$g(x+h) - g(x) = \frac{1}{\sin(x+h)} - \frac{1}{\sin(x)}$$

Find a common denominator:

$$g(x+h) - g(x) = \frac{\sin(x) - \sin(x+h)}{\sin(x+h)\sin(x)}$$

$$g'(x) = \lim_{h \to 0} \frac{\frac{\sin(x) - \sin(x+h)}{\sin(x+h)\sin(x)}}{h}$$ $$g'(x) = \lim_{h \to 0} \frac{\sin(x) - \sin(x+h)}{h \sin(x+h)\sin(x)}$$

Use the sum-to-product identity: $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.
Let $A=x$ and $B=x+h$. Then $A+B = 2x+h$ and $A-B = -h$.

$$\sin(x) - \sin(x+h) = 2 \cos\left(\frac{2x+h}{2}\right) \sin\left(\frac{-h}{2}\right)$$

Using $\sin(-\theta) = -\sin(\theta)$:

$$\sin(x) - \sin(x+h) = -2 \cos\left(x+\frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$

$$g'(x) = \lim_{h \to 0} \frac{-2 \cos\left(x+\frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h \sin(x+h)\sin(x)}$$

Rearrange terms to use the special limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$$g'(x) = \lim_{h \to 0} \left[ \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \cdot \frac{-\cos\left(x+\frac{h}{2}\right)}{\sin(x+h)\sin(x)} \right]$$

As $h \to 0$:
$\lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} = 1$
$\lim_{h \to 0} \cos\left(x+\frac{h}{2}\right) = \cos(x)$
$\lim_{h \to 0} \sin(x+h) = \sin(x)$

$$g'(x) = 1 \cdot \frac{-\cos(x)}{\sin(x)\sin(x)}$$ $$g'(x) = -\frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\sin(x)}$$

Using the identities $\frac{\cos(x)}{\sin(x)} = \cot(x)$ and $\frac{1}{\sin(x)} = \csc(x)$:

$$\boxed{g'(x) = -\cot(x)\csc(x)}$$