Question

pf!! me ajudem!

Se f(x) = $\frac{cosx}{1+senx}$ então o valor de f'(π) é:
a) -1
b) 1
c) 0
d) 1/2
e)2

Answer 1

Resposta:

f'(x)=[-sen(x)*(1+sen(x) )- cos(x)* cos(x)]/ [1+sen(x)]²

f'(π)=[-sen(π)*(1+sen(π)) - cos(π)* cos(π)]/ [1+sen(π)]²

f'(π)=[-0*(1+sen(π)) - 1* 1]/ [1+0]²

f'(π)=-1/ 1

f'(π)=-1

letra A

Answer 2

Oie, tudo bom?

Resposta: a) - 1.

$f(x) = \frac{ \cos(x) }{1 + \sin(x) } \\ f'(x) = \frac{d}{dx} ( \frac{ \cos(x) }{1 + \sin(x) } ) \\ f'(x) = \frac{ \frac{d}{dx}( \cos(x) ) \: . \: ( \sin(x)) - \cos(x) \: . \: \frac{d}{dx} (1 + \sin(x) )}{(1 + \sin(x) ) {}^{2} } \\ f'(x) = \frac{ - \sin(x) \: . \: (1 + \sin(x)) - \cos(x) \cos(x) }{(1 + \sin(x)) {}^{2} } \\ f'(x) = \frac{ - \sin(x) - \sin(x) {}^{2} - \cos(x) {}^{2} }{(1 + \sin(x)) {}^{2} } \\ f'(x) = \frac{ - \sin(x) - 1}{(1 + \sin(x) ){}^{2} } \\ f'(x) = - \frac{ \sin(x) + 1 }{(1 + \sin(x) ) {}^{2} } \\ \boxed{f'(x) = - \frac{1}{1 + \sin(x) } } \\ \boxed{x⟶\pi} \\ f'(\pi) = - \frac{1}{1 + \sin(\pi) } \\ f'(\pi) = - \frac{1}{1 + 0} \\ f'(\pi) = - \frac{1}{1} \\ \boxed{ f'(\pi) = - 1}$

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