f-x-x-ln-x-1-e-x-f-x-

Question Number 4993 by love math last updated on 29/Mar/16

f (x) = \frac{x l n x}{1 + e^{x}}

f^{'} (x) - ?

Answered by FilupSmith last updated on 30/Mar/16

f (x) = \frac{u}{v} \Rightarrow f^{'} (x) = \frac{u^{'} v - {u v}^{'}}{v^{2}}

u = x \ln (x) \Rightarrow u^{'} = 1 + \ln (x)

v = 1 + e^{x} \Rightarrow v^{'} = e^{x}

\Rightarrow v^{2} = {(1 + e^{x})}^{2}

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

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

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

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

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

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

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

Commented by FilupSmith last updated on 30/Mar/16

Correct if wrong