Ответ:

Производная произведения :   [tex]\bf (uv)'=u'v+uv'[/tex]   .

[tex]\bf f(x)=(4x-\pi )(tgx+sinx)\ \ ,\ \ \ x_0=\dfrac{\pi }{4}\\\\u=4x-\pi \ \ ,\ \ v=tgx+sinx\\\\f'(x)=(4x-\pi )'(tgx+sinx)+(4x-\pi )(tgx+sinx)'=\\\\=4\, (tgx+sinx)+(4x-\pi )(\dfrac{1}{cos^2x}+cosx)\\\\f'(\dfrac{\pi }{4})=4\, (1+\dfrac{\sqrt2}{2})+(\pi -\pi )(\dfrac{1}{\frac{1}{2}}+\dfrac{\sqrt2}{2})=4+2\sqrt2[/tex]