Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \frac{{x\sin x}}{{1 + \tan x}}\\
\left( {x\sin x} \right)' = x'.\sin x + x.\left( {\sin x} \right)' = \sin x + x\cos x\\
\left( {1 + \tan x} \right)' = \left( {\tan x} \right)' = \frac{1}{{{{\cos }^2}x}}\\
\Rightarrow y' = \frac{{\left( {x\sin x} \right)'.\left( {1 + \tan x} \right) - \left( {1 + \tan x} \right)'.\left( {x\sin x} \right)}}{{{{\left( {1 + \tan x} \right)}^2}}}\\
= \frac{{\left( {\sin x + x\cos x} \right)\left( {1 + \frac{{\sin x}}{{\cos x}}} \right) - \frac{1}{{{{\cos }^2}x}}.x.\sin x}}{{{{\left( {1 + \frac{{\sin x}}{{\cos x}}} \right)}^2}}}\\
= \frac{{\frac{{\sin x\cos x + x{{\cos }^2}x}}{{\cos x}}.\frac{{\cos x + \sin x}}{{\cos x}} - \frac{{x.\sin x}}{{{{\cos }^2}x}}}}{{{{\left( {\frac{{\sin x + \cos x}}{{\cos x}}} \right)}^2}}}\\
= \frac{{\left( {\sin x\cos x + x.{{\cos }^2}x} \right)\left( {\sin x + \cos x} \right) - x\sin x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}\\
= \frac{{\sin x\cos x + x{{\cos }^2}x}}{{\sin x + \cos x}} - \frac{{x\sin x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}
\end{array}\)
