Question

utilizando as regras de derivação, calcule F(x)´de F(x)=sec(x)+sen(x)/cos(x)

Answer 1

Encontrar a derivada da função

$\mathsf{f(x)=sec\,x+\dfrac{sen\,x}{cos\,x}}\\\\\\ \mathsf{f(x)=\dfrac{1}{cos\,x}+\dfrac{sen\,x}{cos\,x}}\\\\\\ \mathsf{f(x)=\dfrac{1+sen\,x}{cos\,x}}$


Derivamos, usando a Regra do Quociente:

$\mathsf{f'(x)=\left(\dfrac{1+sen\,x}{cos\,x}\right)'}\\\\\\ \mathsf{f'(x)=\dfrac{(1+sen\,x)'\cdot cos\,x-(1+sen\,x)\cdot (cos\,x)'}{(cos\,x)^2}}\\\\\\ \mathsf{f'(x)=\dfrac{(0+cos\,x)\cdot cos\,x-(1+sen\,x)\cdot (-sen\,x)}{cos^2\,x}}\\\\\\ \mathsf{f'(x)=\dfrac{cos^2\,x-(-sen\,x-sen^2\,x)}{cos^2\,x}}\\\\\\ \mathsf{f'(x)=\dfrac{cos^2\,x+sen\,x+sen^2\,x}{cos^2\,x}}$

$\mathsf{f'(x)=\dfrac{(cos^2\,x+sen^2\,x)+sen\,x}{cos^2\,x}}\\\\\\ \mathsf{f'(x)=\dfrac{1+sen\,x}{cos^2\,x}}\\\\\\ \mathsf{f'(x)=\dfrac{1}{cos^2\,x}+\dfrac{sen\,x}{cos^2\,x}}\\\\\\ \mathsf{f'(x)=\left(\dfrac{1}{cos\,x}\right)^{\!2}+\dfrac{sen\,x}{cos\,x}\cdot \dfrac{1}{cos\,x}}\\\\\\ \boxed{\begin{array}{c}\mathsf{f'(x)=sec^2\,x+tg\,x\cdot\,sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


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Tags: derivada função trigonométrica quociente secante sec tangente tg seno sen cosseno cos cálculo diferencial