Question

determine a derivada:

f(x) $\frac{secx}{1+tgx}$

Answer 1

Primeiro, lembremos das derivadas das funções trigonométricas secante e tangente de $x:$

$\dfrac{d}{dx}(\sec x)=\sec x\cdot \mathrm{tg\,}x\\\\\\ \dfrac{d}{dx}(\mathrm{tg\,} x)=\sec^2 x$
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Dada a função

$f(x)=\dfrac{\sec x}{1+\mathrm{tg\,}x}$

calcularemos sua derivada usando a Regra do Quociente:

$\dfrac{df}{dx}(x)=\dfrac{\frac{d}{dx}(\sec x)\cdot (1+\mathrm{tg\,}x)-\sec x\cdot \frac{d}{dx}(1+\mathrm{tg\,}x)}{(1+\mathrm{tg\,}x)^2}\\\\\\ \dfrac{df}{dx}(x)=\dfrac{(\sec x\cdot \mathrm{tg\,}x)\cdot (1+\mathrm{tg\,}x)-\sec x\cdot (0+\sec^2 x)}{(1+\mathrm{tg\,}x)^2}\\\\\\ \boxed{\begin{array}{c}\dfrac{df}{dx}(x)=\dfrac{\sec x\cdot \mathrm{tg\,}x\cdot (1+\mathrm{tg\,}x)-\sec^3 x}{(1+\mathrm{tg\,}x)^2} \end{array}}$