Question

Даю 44 балла, какие получатся, пожалуйста!Производная!!!!!

Answer 1

$1)f(x)=(x-5)*\frac{1}{x}\\\\f'(x)=(x-5)'*\frac{1}{x}+(x-5)*(\frac{1}{x})'=1*\frac{1}{x}+(x+5)*(-\frac{1}{x^{2}})} =\frac{1}{x}-\frac{x+5}{x^{2}}=\frac{x-x-5}{x^{2}}=-\frac{5}{x^{2}}$

$2)f(x)=\frac{x^{3}-12}{x+4}\\\\f'(x)=\frac{(x^{3}-12)'*(x+4)-(x^{3}-12)*(x+4)'}{(x+4)^{2}} =\frac{3x^{2}*(x+4)-(x^{3}-12)}{(x+4)^{2}}=\frac{3x^{3}+12x^{2}-x^{3} +12}{(x+4)^{2}}=\frac{2x^{3}+12x^{2}+12  }{(x+4)^{2} }$

$3)f(x)=\frac{2}{x}*(x+3)\\\\f'(x)=(\frac{2}{x})'*(x+3)+\frac{2}{x}*(x+3)'=-\frac{2}{x^{2}}*(x+3)+\frac{2}{x}=\frac{-2x-6+2x}{x^{2}}=-\frac{6}{x^{2}}$

$4)f(x)=\frac{x^{2}+10}{x-5}\\\\f'(x)=\frac{(x^{2}+10)'*(x-5)-(x^{2} +10)*(x-5)' }{(x-5)^{2}}=\frac{2x*(x-5)-(x^{2}+10)}{(x-5)^{2}}=\frac{2x^{2}-10x-x^{2}-10}{(x-5)^{2}}=\frac{x^{2}-10x-10}{(x-5)^{2} }$

$5)f(x)=(3x-1)*\sqrt{x} \\\\f'(x)=(3x-1)'*\sqrt{x}+(3x-1)*(\sqrt{x} )'=3\sqrt{x}+(3x-1)*\frac{1}{2\sqrt{x}}=3\sqrt{x}+\frac{3x-1}{2\sqrt{x}}=\frac{6x+3x-1}{2\sqrt{x}}=\frac{9x-1}{2\sqrt{x}}$

$6)f(x)=\frac{3x^{2} -1}{x+2}\\\\f'(x)=\frac{(3x^{2}-1)'*(x+2)-(3x^{2}-1)*(x+2)'}{(x+2)^{2}}=\frac{6x*(x+2)-(3x^{2}-1)}{(x+2)^{2}}=\frac{6x^{2}+12x-3x^{2}+1}{(x+2)^{2}} =\frac{3x^{2}+12x+1 }{(x+2)^{2}}$