Question

Qual a derivada de y=(3sec²x)/x

Answer 1

$y = \displaystyle \frac{3 \sec^{2}(x)}{x} \\ \\ \\ y' = \frac{[3 \sec^{2}(x)]' \cdot x - [3 \sec^{2}(x)] \cdot x'}{x^{2}} \\ \\ \\ y' = \frac{3 \cdot 2 \cdot \sec^{2-1}(x) \cdot \sec(x)' \cdot x - 3 \sec^{2}(x)}{x^{2}} \\ \\ \\ y' = \frac{6 \sec(x) \cdot \tan(x) \cdot \sec(x) \cdot x-3 \sec^{2}(x)}{x^{2}} \\ \\ \\ y' = \frac{6 \cdot \displaystyle \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos^{2}(x)} \cdot x - 3 \cdot (\frac{1}{\cos(x)})^{2}}{x^{2}}$

$y' = \displaystyle \frac{\displaystyle \frac{6x \sin(x)}{\cos^{3}(x)}-\frac{3}{\cos^{2}(x)}}{x^{2}} \\ \\ \\ y' = \frac{\displaystyle \frac{6x \cos^{2}(x) \sin(x)- 3 \cos^{3}(x)}{\cos^{5}(x)}}{x^{2}} \\ \\ \\ y' = \displaystyle \frac{6x \cos^{2}(x) \sin(x)- 3 \cos^{3}(x)}{\cos^{5}(x)} \cdot \frac{1}{x^{2}} \\ \\ \\ y' = \displaystyle \frac{6x \cos^{2}(x) \sin(x)- 3 \cos^{3}(x)}{x^{2} \cos^{5}(x)}$

$y' = \displaystyle \frac{\cos^{2}(x) [ 6x \sin(x)- 3 \cos(x)]}{x^{2} \cos^{5}(x)} \\ \\ \\ y' = \frac{6x \sin(x)-3 \cos(x)}{x^{2} \cos^{3}(x)}$