Rozwiązanie:
1) [tex]f(x)=\frac{x^{6}+tgx}{lnx}[/tex]
[tex]f'(x)=\frac{(6x^{5}+\frac{1}{cos^{2}x})lnx-\frac{1}{x}(x^{6}+tgx) }{ln^{2}x} =\frac{6x^{5}lnx+\frac{lnx}{cos^{2}x} -x^{5}-\frac{tgx}{x} }{ln^{2}x} =\frac{x^6(6lnx-1)-tgx+xlnxsec^{2}x}{xln^{2}x}[/tex]
2) [tex]g(x)=(x^{6}-x^{5})^{tgx}=e^{ln(x^6-x^5)^{tgx}}=e^{tgxln(x^6-x^5)[/tex]
[tex]g'(x)=e^{tgxln(x^6-x^5)}*(tgxln(x^6-x^5))'=e^{tgxln(x^6-x^5)}*(\frac{ln(x^6-x^5)}{cos^2x} +\frac{6x^5-5x^4}{x^6-x^5} *tgx)=e^{tgxln(x^6-x^5)}(\frac{ln(x^6-x^5)}{cos^2x}+\frac{6x^5tgx-5x^4tgx}{x^6-x^5})=(x^6-x^5)^{tgx}(\frac{ln(x^6-x^5)}{cos^2x}+\frac{tgx(6x^5-5x^4)}{x^6-x^5})[/tex]
