Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\sqrt x \,\,\,khi\,\,x > 1\\{x^2}\,\,\,\,\,khi\,\,x \le 1\end{array} \right.\). Tính \(f'\left( 1 \right)\) ?

\(y'' = \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}\) 

\(y'' = \dfrac{1}{{\sqrt {2x + 5} }}\) 

\(y'' =  - \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}\) 

$y''=-\dfrac{1}{\sqrt{2x+5}}$

\(y'\left( 1 \right) =  - 4\).

\(y'\left( 1 \right) =  - 5\).

\(y'\left( 1 \right) =  - 3\).

\(y'\left( 1 \right) =  - 2\).

\(y' = \dfrac{{ - 13{x^2} - 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\)          

\(y' = \dfrac{{ - 13{x^2} + 5x + 11}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\) 

\(y' = \dfrac{{ - 13{x^2} + 5x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\) 

\(y' = \dfrac{{ - 13{x^2} + 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\)

\({y^{(5)}} =  - \dfrac{{120}}{{{{(x + 1)}^6}}}\).

\({y^{(5)}} = \dfrac{{120}}{{{{(x + 1)}^6}}}\).

\({y^{(5)}} = \dfrac{1}{{{{(x + 1)}^6}}}\).

\({y^{(5)}} =  - \dfrac{1}{{{{(x + 1)}^6}}}\).

\(y' = \sin \left( {x + \dfrac{\pi }{2}} \right)\) 

\(y'' = \sin \left( {x + \pi } \right)\) 

\(y''' = \sin \left( {x + \dfrac{{3\pi }}{2}} \right)\) 

\({y^{\left( 4 \right)}} = \sin \left( {2\pi  - x} \right)\) 

\(y = 7x + 2\).

\(y = 7x - 2\).

\(y =  - 7x + 2\) .

\(y =  - 7x - 2\).

\(x = \dfrac{\pi }{2}\) 

\(x = 0\) hoặc \(x = \dfrac{\pi }{6}\) 

\(x = 0\) hoặc \(x = \dfrac{\pi }{3}\) 

\(x = 0\) hoặc \(x = \dfrac{\pi }{2}\)