\(f'\left( x \right) = \frac{1}{{{{\cos }^2}x}} \Rightarrow f'\left( { - \frac{\pi }{6}} \right) = \frac{1}{{{{\cos }^2}\left( { - \frac{\pi }{6}} \right)}} = \frac{4}{3}\)

\(f'\left( x \right) =  - \frac{1}{{{{\sin }^2}x}} \Rightarrow f'\left( { - \frac{\pi }{3}} \right) =  - \frac{1}{{{{\sin }^2}\left( { - \frac{\pi }{3}} \right)}} =  - \frac{4}{3}\)

\(f'\left( x \right) = {10^x}.\ln 10 \Rightarrow f'\left( { - 1} \right) = {10^{ - 1}}.\ln 10 = \frac{{\ln 10}}{{10}}\)

a,

\(y' = 6x - 4 \Rightarrow y'' = 6\)

Tại \({x_0} =  - 2 \Rightarrow y''( - 2) = 6\)

b,

\(\begin{array}{l}y' = \frac{2}{{\left( {2x + 1} \right)\ln 3}}\\ \Rightarrow y'' = \left( {2.\frac{1}{{\left( {\left( {2x + 1} \right)\ln 3} \right)}}} \right)' =  - 2.\frac{{\left( {\left( {2x + 1} \right)\ln 3} \right)'}}{{{{\left( {\left( {2x + 1} \right)\ln 3} \right)}^2}}}\\ =  - 2\frac{{2\ln 3}}{{{{\left( {\left( {2x + 1} \right)\ln 3} \right)}^2}}} = \frac{{ - 4\ln 3}}{{{{\left( {\left( {2x + 1} \right)\ln 3} \right)}^2}}}\end{array}\)

Tại \({x_0} = 3 \Rightarrow y''(3) = \frac{{ - 4\ln 3}}{{{{\left( {\left( {2.3 + 1} \right)\ln 3} \right)}^2}}} = \frac{{ - 4\ln 3}}{{{{\left( {7\ln 3} \right)}^2}}} = \frac{{ - 4}}{{49\ln 3}}\)

c, \(y' = 4{e^{4x + 3}} \Rightarrow y'' = 16{e^{4x + 3}}\)

Tại \({x_0} = 1 \Rightarrow y''(1) = 16.{e^{4.1 + 3}} = 16.{e^7}\)

d,

\(y' = 2\cos \left( {2x + \frac{\pi }{3}} \right) \Rightarrow y'' =  - 4\sin \left( {2x + \frac{\pi }{3}} \right)\)

Tại \({x_0} = \frac{\pi }{6} \Rightarrow y''\left( {\frac{\pi }{6}} \right) =  - 4\sin \left( {2.\frac{\pi }{6} + \frac{\pi }{3}} \right) =  - 2\sqrt 3 \)

e,

\(y' =  - 3.\sin \left( {3x - \frac{\pi }{6}} \right) \Rightarrow y'' =  - 9.\cos \left( {3x - \frac{\pi }{6}} \right)\)

Tại \({x_0} = 0 \Rightarrow y''(0) =  - 9.\cos \left( {3.0 - \frac{\pi }{6}} \right) = \frac{{ - 9\sqrt 3 }}{2}\)

Ta có: \(\mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = f'\left( {{x_0}} \right);\mathop {\lim }\limits_{x \to {x_0}} \frac{{g\left( x \right) - g\left( {{x_0}} \right)}}{{x - {x_0}}} = g'\left( {{x_0}} \right)\)

Vậy \(h'\left( {{x_0}} \right) = f'\left( {{x_0}} \right) + g'\left( {{x_0}} \right)\).

\(\begin{array}{l}f'\left( x \right) = {\left( {\sqrt x } \right)'} = \frac{1}{{2\sqrt x }}\\ \Rightarrow f'\left( 9 \right) = \frac{1}{{2\sqrt 9 }} = \frac{1}{{2.3}} = \frac{1}{6}\end{array}\)

1) \(f\left(x\right)=2x-5\)

\(f'\left(x\right)=2\)

\(\Rightarrow f'\left(4\right)=2\)

2) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)

\(\Rightarrow y'=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)

3) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)

\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)

\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)

a: \(f'\left(x_0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x0\right)}{x-x0}=\lim\limits_{x\rightarrow x0}\dfrac{c-c}{x-x0}=0\)

b: \(f'\left(x0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x0\right)}{x-x0}=\lim\limits_{x\rightarrow x0}\dfrac{x-x0}{x-x0}=1\)

f(x) = sin x

=> f'(x) = cos x

\(f'\left(\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}\right)=0\)