Question

Найти производную функции $f(x)=(frac{tg^{3}(3x+5)e^{x^{2}} 2^{x}}{sqrt{x}(cosx)^{tgx} } )$

Answer 1

$f(x)=(frac{tg^{3}(3x+5)e^{x^{2}}2^{x}}{sqrt{x}(cosx)^{tgx}});\\f'(x)=frac{(tg^{3}(3x+5)e^{x^{2}}2^{x})'*(sqrt{x}(cosx)^{tgx})-(tg^{3}(3x+5)e^{x^{2}}2^{x})*(sqrt{x}(cosx)^{tgx})'}{(sqrt{x}(cosx)^{tgx})^2};\\(tg^{3}(3x+5)e^{x^{2}}2^{x})'=(tg^{3}(3x+5))'*(e^{x^2}*2^x)+(tg^{3}(3x+5))*(e^{x^2}*2^x)';\\(tg^{3}(3x+5))'=3tg^{2}(3x+5)*(tg(3x+5))'=frac{3tg^{2}(3x+5)}{cos^2{(3x+5)}}*(3x+5)'=\(frac{3tg(3x+5)}{cos{(3x+5)}})^2;$

$(e^{x^2}*2^x)'=(e^{x^2})'*2^x+e^{x^2}*(2^x)'=e^{x^2}*(x^2)'*2^x+e^{x^2}*2^{x}*ln{2}=\\e^{x^2}*2^x(2x+ln{2});\\(tg^{3}(3x+5)e^{x^{2}}2^{x})'=(frac{3tg(3x+5)}{cos{(3x+5)}})^2*(e^{x^2}*2^x)+(tg^{3}(3x+5))(e^{x^2}*2^x(2x+ln{2}))=\\(e^{x^2}*2^x)(tg^{2}(3x+5))(frac{9}{cos^2{(3x+5)}}+(tg(3x+5))(2x+ln{2}))=\\frac{e^{x^2}*2^x*tg^{2}(3x+5)}{cos^2{(3x+5)}}(9+frac{1}{2}sin{(6x+10)}(2x+ln{2}));$

$(sqrt{x}(cosx)^{tgx})'=frac{(cosx)^{tgx}}{2sqrt{x}}+sqrt{x}*((cosx)^{tgx})';\\((cosx)^{tgx})'=(cosx)^{tgx}*(tg(x)*ln{(cos{x})})';\\(tg(x)*ln{(cos{x})})'=frac{ln{(cos{x})}}{cos^2{x}}-frac{tg(x)*sin{x}}{cos{x}}=frac{ln{(cos{x})}-sin^2{x}}{cos^2{x}};\\(sqrt{x}(cosx)^{tgx})'=frac{(cosx)^{tgx}}{2sqrt{x}}(1+2x*frac{ln{(cos{x})}-sin^2{x}}{cos^2{x}});$

$f'(x)=(frac{e^{x^2}*2^x*tg^{2}(3x+5)}{cos^2{(3x+5)}}(9+frac{1}{2}sin{(6x+10)}(2x+ln{2}))*(sqrt{x}(cosx)^{tgx})-\\(tg^{3}(3x+5)e^{x^{2}}2^{x})*frac{(cosx)^{tgx}}{2sqrt{x}}(1+2x*frac{ln{(cos{x})}-sin^2{x}}{cos^2{x}})):(x*(cosx)^{2tgx})=\\(frac{e^{x^2}*2^x*tg^{2}(3x+5)}{cos^2{(3x+5)}}(9+frac{1}{2}sin{(6x+10)}(2x+ln{2}))*(sqrt{x})-(tg^{3}(3x+5)e^{x^{2}}2^{x})*\\frac{1}{2sqrt{x}}(1+2x*frac{ln{(cos{x})}-sin^2{x}}{cos^2{x}})):(x*(cosx)^{tgx})$