Question

Derivada de
$y= \frac{3arctgx}{x^4}$

Answer 1

$y=\dfrac{3\,\mathrm{arctg\,}x}{x^{4}}$

Derivando pela regra do quociente:

$\dfrac{dy}{dx}=\dfrac{\frac{d}{dx}(3\,\mathrm{arctg\,}x)\cdot x^{4}-3\,\mathrm{arctg\,}x\cdot \frac{d}{dx}(x^{4})}{(x^{4})^{2}}\\\\\\ \dfrac{dy}{dx}=\dfrac{3\cdot \frac{d}{dx}(\mathrm{arctg\,}x)\cdot x^{4}-3\,\mathrm{arctg\,}x\cdot 4x^{3}}{x^{8}}\\\\\\ \dfrac{dy}{dx}=\dfrac{3\cdot \frac{1}{1+x^{2}}\cdot x^{4}-12x^{3}\,\mathrm{arctg\,}x}{x^{8}}\\\\\\ \dfrac{dy}{dx}=\dfrac{3x^{4}\cdot \frac{1}{1+x^{2}}-12x^{3}\,\mathrm{arctg\,}x}{x^{8}}\\\\\\ \dfrac{dy}{dx}=\dfrac{\diagup\!\!\!\!\! x^{3}\cdot \left(3x\cdot \frac{1}{1+x^{2}}-12\,\mathrm{arctg\,}x \right)}{\diagup\!\!\!\!\! x^{3}\cdot x^{5}}$

$\dfrac{dy}{dx}=\dfrac{3x\cdot \frac{1}{1+x^{2}}-12\,\mathrm{arctg\,}x}{x^{5}}\\\\\\ \dfrac{dy}{dx}=\dfrac{3x\cdot \frac{1}{1+x^{2}}-12\,\mathrm{arctg\,}x}{x^{5}}\cdot \dfrac{1+x^{2}}{1+x^{2}}\\\\\\ \therefore~~\boxed{\begin{array}{c} \dfrac{dy}{dx}=\dfrac{3x-12\,(1+x^{2})\,\mathrm{arctg\,}x}{x^{5}\,(1+x^{2})} \end{array}}$