Question

Utilizando a regra de derivação, calcule y’ em cada caso:



(A) y = cotg x



(B) y = sec x

Answer 1

A regra de derivação a ser utilizada é a regra da derivada do quociente:

$\left(\dfrac{u}{v} \right )'=\dfrac{u'v-uv'}{v^{2}}$


(A) $y=\mathrm{cotg\,}x$

$y=\dfrac{\cos x}{\mathrm{sen\,}x}\\ \\ \\ y'=\left(\dfrac{\cos x}{\mathrm{sen\,}x} \right )'\\ \\ y'=\dfrac{\left(\cos x \right )'\cdot \mathrm{sen\,}x-\cos x\cdot \left(\mathrm{sen\,}x \right )'}{\left(\mathrm{sen\,}x \right )^{2}}\\ \\ y'=\dfrac{\left(-\mathrm{sen\,}x \right )\cdot \mathrm{sen\,}x-\cos x\cdot \left(\cos x \right )}{\mathrm{sen^{2}\,}x}\\ \\ y'=\dfrac{-\mathrm{sen^{2}\,}x-\cos^{2} x}{\mathrm{sen^{2}\,}x}\\ \\ y'=\dfrac{-\left(\mathrm{sen^{2}\,}x+\cos^{2} x \right )}{\mathrm{sen^{2}\,}x}\\ \\ y=\dfrac{-1}{\mathrm{sen^{2}}x}\\ \\ y=-\left(\dfrac{1}{\mathrm{sen\,}x} \right )^{2}\\ \\ y=-\mathrm{cossec^{2}\,}x$


(B) $y=\sec x$

$y=\dfrac{1}{\cos x}\\ \\ \\ y'=\dfrac{\left(1 \right )'\cdot \cos x-1\cdot \left(\cos x \right )'}{\left(\cos x \right )^{2}}\\ \\ y'=\dfrac{0\cdot \cos x-1\cdot \left(-\mathrm{sen\,}x \right )}{\cos^{2} x}\\ \\ y'=\dfrac{\mathrm{sen\,}x}{\cos^{2} x}\\ \\ y'=\dfrac{\mathrm{sen\,}x}{\cos x}\cdot \dfrac{1}{\cos x}\\ \\ y'=\mathrm{tg\,}x\cdot \sec x$