(no subject)
4. Evaluate $f'(x)$ for the following functions: \\
a. $\frac {6x - 3x^2}{2x^2}$
$$f(x) = 6x - 3x^3 $$
$$f'(x) = 6 - 9x^2$$
$$g(x) = 2x^2$$
$$g'(x) = 4x$$
$$\frac {(6 - 9x^2)(2x^2) - (6x - 3x^3)(4x)}{(2x^2)^2}$$
$$\frac {(12x^2 - 18x^4) - (24x^2 - 12x^4)}{(2x^2)^2}$$
$$\frac {-12x^2 - 6x^4}{(2x^2)^2}$$
$$\frac {-3x^2 - 6}{2x^2}$$
\\
b.$(3x - 7)(2x^2 - 4x + 3)$ $$g(x) = 3x - 7$$ $$g'(x) = 3$$ $$h(x) = 2x^2 - 4x +3$$ $$h'(x) = 4x - 4$$ $$(3)(2x^2 - 4x + 3) + (4x - 4)(3x - 7)$$ $$6x^2 - 12x + 9 + 12x^2 - 28x - 12x + 28$$ $$18x^2 -52x + 37$$\\
c. $(2a + x)(x^2 - b)$ $$g(x) = 2a + x$$ $$g'(x) = 1$$ $$h(x) = x^2 - b$$ $$h'(x) = 2x$$ $$(2x)(2a + x) = (1)(x^2 -b)$$ $$ 4ax + 2x^2 + x^2 - b$$ $$ 3x^2 + 4ax - b$$\\
d. $\frac {\sqrt [2] {x^3}} {2x^2}$
$$f(x) = \sqrt [2] {x^3}$$
$$f'(x) = \frac {3}{2}x^\frac {1}{2}$$
$$g(x) = 2x^2$$
$$g'(x) = 4x$$
$$\frac {f'(x)g(x) - f(x)g'(x)}{g^2(x)}$$
$$\frac {(\frac {3}{2}x^\frac {1}{2})(2x^2) - (x^\frac {3}{2})(4x)}{4x^4}$$
$$\frac {-x^\frac{5}{2}}{4x^4}$$
$$\frac {-1}{4\sqrt {x^3}}
\\
e. $ln \sqrt {1 - x^2}$
$$f(x) = ln\sqrt {1 - x^2}$$
$$g(x) = \sqrt {1 - x^2}$$
$$h(x) = 1 - x^2$$
$$[\frac {1}{\sqrt {1 - x^2}}][\frac {1}{2}(1 - x^2)^\frac {-1}{2}][-2x]$$
$$[\frac {1}{\sqrt {1 - x^2}}][\frac {1}{2\sqrt {1 - x^2}}][-2x]$$
$$\frac {-2x}{2(1 - x^2)}$$
$$\frac {-x}{1 - x^2}$$
\\
f. $e^x ln x$
$$f'(x) = (e^x)(ln x + e^x)(\frac {1}{x})$$
$$(e^x)(ln x) + \frac {e^x}{x}$$
\\
g. $e^{x^3} ln x^2$
$$e^{x^3} 2ln x$$
$$f(x) = e^{x^3}$$
$$f'(x) = (e^{x^3})(3x^2)$$
$$g(x) = 2 ln x$$
$$g'(x) = \frac {2}{x} $$
$$[(e^{x^3})(3x^2)][2 ln x] + [e^{x^3}][\frac {2}{x}] $$
$$[e^{x^3}][3x^2][2 ln x] + \frac {2e^{x^3}}{x} $$
\\
h. $ln(e^x + 1) $
$$(\frac {1}{e^x + 1})(e^x) $$
$$\frac {e^x}{e^x + 1}$$
\\