Tính đạo hàm của các hàm số sau:a) \(y = \frac{{{x^{10}}}}{5} - 2{x^4} + 2010\)b) \(y = \frac{{2x - 5}}{{1 - x}}\)c) \

* Hướng dẫn giải

a) \(y' = {\left( {\frac{{{x^{10}}}}{5} - 2{x^4} + 2010} \right)'} = 2{x^5} - 8x\)

b) \(y' = {\left( {\frac{{2x - 5}}{{1 - x}}} \right)'} = \frac{{(2x - 5)'(1 - x) - (2x - 5)(1 - x)'}}{{{{(1 - x)}^2}}} = \frac{{ - 3}}{{{{(1 - x)}^2}}}\)

c) \(y' = {\left( {\sqrt {{x^3} - x + 10} } \right)'} = \frac{{({x^3} - x + 10)'}}{{2\sqrt {{x^3} - x + 10} }} = \frac{{3{x^2} - 1}}{{2\sqrt {{x^3} - x + 10} }}\)

d) \(y' = {\left[ {{{({x^2} - 10x + 1)}^{10}}} \right]'} = 10{({x^2} - 10x + 1)^9}({x^2} - 10x + 1)' = 10{({x^2} - 10x + 1)^9}(2x - 10)\)

e) \(y = (\sin 2x - 10\cos x)' = (\sin 2x)' - 10(\cos x)' = 2\cos 2x + 10\sin x\)

f) \(y' = {\left[ {\sqrt {1 - {{\cot }^2}\frac{x}{2}} } \right]'} = \frac{{{{\left( {1 - {{\cot }^2}\frac{x}{2}} \right)}'}}}{{2\sqrt {1 - {{\cot }^2}\frac{x}{2}} }} = \frac{{ - 2\cot \frac{x}{2}.{{\left( {\cot \frac{x}{2}} \right)}'}}}{{2\sqrt {1 - {{\cot }^2}\frac{x}{2}} }} = \frac{{\cot \frac{x}{2}.\frac{{{{\left( {\frac{x}{2}} \right)}'}}}{{{{\sin }^2}\frac{x}{2}}}}}{{\sqrt {1 - {{\cot }^2}\frac{x}{2}} }} = \frac{{\cot \frac{x}{2}}}{{2{{\sin }^2}\frac{x}{2}.\sqrt {1 - {{\cot }^2}\frac{x}{2}} }}\)