Rozwiązanie:
[tex]a)[/tex]
[tex]f(x)=\frac{\sqrt[3]{x^{2}} }{cos5x} =\frac{x^{\frac{2}{3} }}{cos5x} \\f'(x)=\frac{\frac{2}{3} x^{-\frac{1}{3}}\cdot cos5x +5sin5x \cdot x^{\frac{2}{3} }}{cos^{2}5x}[/tex]
[tex]b)[/tex]
[tex]f(x)=cos(sin(tgx))\\f'(x)=-sin(sin(tgx)) \cdot (sin(tgx))'=-sin(sin(tgx)) \cdot cos(tgx) \cdot (tgx)'=-sin(sin(tgx)) \cdot cos(tgx) \cdot \frac{1}{cos^{2}x} =-\frac{sin(sin(tgx))cos(tgx)}{cos^{2}x}[/tex]
