Question

Considere as funções f(x) = lnx/e^x e g(x) = ( ln x )^3 Calcule a derivada da soma f(x) + g(x) no ponto x = 1.

Answer 1

derivadas:

$\bmatrix( u^{N} )'= N*u^{N-1}*u'\\\\(e^u)'= e^u *u' \\\\(ln(u))' = \frac{1}{u}*u' \\\\( \frac{U}{V} )'= \frac{U'*V-U*V'}{V^2} \end$


a derivada de uma soma é a soma das derivadas então:

$(f(x)+g(x))' = f'(x)+g'(x) \\ \text{em x=1}\\\\ \boxed{(f(1)+g(1))' = f'(1)+g'(1) }$



temos:

$f(x)= \frac{ln(x)}{e^x} \\\\f'(x)= \frac{(ln(x))'*e^x - ln(x)*(e^x)'}{(e^x)^2} \\\\f'(x)= \frac{ \frac{1}{x}*e^x-ln(x)*e^x }{(e^x)^2} \\\\f'(1)= \frac{1*e^1-ln(1)*e^1}{(e^1)^2} = \frac{e}{e^2} = \frac{1}{e}$


$g(x)=(ln(x))^3\\\\g'(x)=3*(ln(x))^{3-1}*(ln(x))'\\\\g'(x)=3(ln(x))^2 * \frac{1}{x} \\\\g'(1)=3*(ln(1))^2* \frac{1}{1} =0$

a derivada da soma em x=1
$\boxed{\boxed{(f(1)+g(1))' = \frac{1}{e}+0 = \frac{1}{e} }}$