Предмет: Математика, автор: Ад1

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

Ответы

Автор ответа: WhatYouNeed

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

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

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

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