Question

Com a regra da cadeia, determine f'(x) sendo f(x) = $\sqrt{x+senx$

Answer 1

Resposta:

Explicação passo-a-passo:

Temos que

$f(x)=\sqrt{x+senx}=>f(x)=(x+senx)^{\frac{1}{2}}$

Logo, fazendo

u = x + senx

$\frac{d}{du}=1+cosxdx$

$f(x)'=\frac{d}{dx}[x^{\frac{1}{2}}].\frac{d}{du}[u]$

$f(x)'=\frac{1}{2}.[x^{\frac{-1}{2}].[1+cosx].$

$f(x)'=\frac{1+cosx}{2.(x+sex)^{\frac{1}{2}}}$

$f(x)'=\frac{1+cosx}{2\sqrt{x+senx}}$

Answer 2

Oie, tudo bom?

$f(x) = \sqrt{x + \sin(x) } \\ f'(x) = \frac{d}{dx} ( \sqrt{x + \sin(x) } ) \\ f'(x) = \frac{d}{dg} ( \sqrt{g} ) \: . \: \frac{d}{dx} (x + \sin(x) ) \\ f'(x) = \frac{1}{2 \sqrt{g} } \: . \: \frac{d}{dx} (x) + \frac{d}{dx} ( \sin(x) ) \\ f'(x) = \frac{1}{2 \sqrt{g} } \: . \: 1 + \cos(x) \\ f'(x) = \frac{1}{2 \sqrt{x + \sin(1) } } \: . \: (1 + \cos(x) ) \\ f'(x) = \frac{1(1 + \cos(x)) }{2 \sqrt{x + \sin(x) } } \\ \boxed{ f'(x) = \frac{1 + \cos(x) }{2 \sqrt{x + \sin(x) } } }$

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