Question

Cálculo I, derivada (imagem)

Answer 1

Explicação passo-a-passo:

Vamos calcular essas derivadas com a Regra do Produto com 3 termos:

$f(x)=u.v.w\implies f'(x)=u'.v.w+u.v'.w+u.v.w'$.

a)

$f(x)=x.e^{x}.\cos(x) \\\\f'(x)=\dfrac{d}{dx} (x).e^{x}.\cos(x)+x.\dfrac{d}{dx} (e^{x}).\cos(x)+x.e^{x}.\dfrac{d}{dx} (\cos(x))\\\\f'(x)=1.e^{x}.\cos(x)+x.e^{x}.\cos(x)+x.e^{x}.(-sen(x))\\\\f'(x)=e^{x}.\cos(x)+x.e^{x}.\cos(x)-x.e^{x}.sen(x)\\\\\boxed{\boxed{f'(x)=e^{x}.(\cos(x)+x.\cos(x)-x.sen(x))}}$

b)

$f(x)=x.\cos(x).(1+\ln(x))\\\\f'(x)=\frac{d}{dx} (x).\cos(x).(1+\ln(x))+x.\frac{d}{dx}(\cos(x)).(1+\ln(x))+x.\cos(x).\frac{d}{dx}(1+\ln(x))\\\\f'(x)=1.\cos(x).(1+\ln(x))+x.(-sen(x)).(1+\ln(x))+x.\cos(x).\dfrac{1}{x}\\\\\boxed{\boxed{f'(x)=\cos(x).(1+\ln(x))-x.sen(x).(1+\ln(x))+\cos(x)}}$