Question

найти производную функции у=5/х; у=3-5х; у=8 √х+0,5соsx; у=sinх/х; у=хсtqx; у=(5х+1)^2

Answer 1

$y'=(frac{5}{x})'=-frac{5}{x^2}\ \ y'=(3-5x)'=3'-5x'=0-5=-5\ \ y'=(8sqrt{x}+0,5cosx)'=0.8*0.5*frac{1}{sqrt{x} } +0.5(cos)x'=frac{0.4}{sqrt{x}} -0.5sinx\\y'=(frac{sinx}{x})'=frac{(sinx)'*x-sinx*x'}{x^2}=frac{x*cosx-sinx}{x^2}\ \y'=(x*ctqx)'=x'ctgx+x*(ctgx)'=ctgx-frac{x}{sin^2x}\ \y'=((5x+1)^2)'=2(5x+1)*(5x+1)'=50x+10$