a) \(f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} x = 1\)

Vậy \(f'\left( 1 \right) = 1\)

b) \(f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to  - 1} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{ - {x^3} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{ - \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \left( {{x^2} - x + 1} \right) = 3\)

Vậy \(f'\left( { - 1} \right) = 3\)

tham khảo:

y′(x0)=\(lim_{x\rightarrow x_0}\)\(\dfrac{f\left(x\right)-f\left(x_0\right)}{x-x_0}\)

=\(lim_{x\rightarrow x_0}\)\(\dfrac{\sqrt{x}-\sqrt{x_0}}{\left(\sqrt{x}-\sqrt{x_0}\right).\left(\sqrt{x}+\sqrt{x_0}\right)}\)

=\(lim_{x\rightarrow x_0}\)\(\dfrac{1}{\sqrt{x}+\sqrt{x_0}}\)

=\(\dfrac{1}{\sqrt{x}+\sqrt{x_0}}\)\(=\dfrac{1}{2\sqrt{x_0}}\)

\(\begin{array}{l}f'({x_0}) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f({x_0})}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^{\frac{1}{2}}} - x_0^{\frac{1}{2}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^{\frac{1}{2}.\ln x}} - {e^{\frac{1}{2}.\ln {x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^{\frac{1}{2}.\ln {x_0}}}.\left( {{e^{\frac{1}{2}\ln x - \frac{1}{2}\ln {x_0}}} - 1} \right)}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x_0^{\frac{1}{2}}\left( {{e^{\frac{1}{2}.\ln x - \frac{1}{2}\ln {x_0}}} - 1} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x_0^{\frac{1}{2}}\left( {\frac{1}{2}\ln x - \frac{1}{2}\ln {x_0}} \right)}}{{x - {x_0}}} = \frac{1}{2}x_0^{\frac{1}{2}}\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \left( {\frac{x}{{{x_0}}}} \right)}}{{x - {x_0}}} = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \left( {1 + \frac{x}{{{x_0}}} - 1} \right)}}{{x - {x_0}}}\\ = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{x}{{{x_0}}} - 1}}{{x - {x_0}}} = \frac{1}{2}x_0^{\frac{1}{2}}\mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{x - {x_0}}}{{{x_0}}}}}{{x - {x_0}}} = \frac{1}{2}x_0^{\frac{1}{2}}\mathop {\lim }\limits_{x \to {x_0}} \frac{1}{{{x_0}}} = \frac{1}{2}x_0^{\frac{1}{2}}.\frac{1}{{{x_0}}}\\ \Rightarrow f'\left( 1 \right) = \frac{1}{2}{.1^{\frac{1}{2}}}.1 = \frac{1}{2}\end{array}\)

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

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) Giả sử ∆x là số gia của số đối tại x₀= 1. Ta có:

∆y = f(1 + ∆x) - f(1) = 7 + (1 + ∆x) - (1 + ∆x)² - (7 + 1 - 1²) = -(∆x)² - ∆x ;

= - ∆x - 1 ; = (- ∆x - 1) = -1.

Vậy f'(1) = -1.

b) Giả sử ∆x là số gia của số đối tại x₀= 2. Ta có:

∆y = f(2 + ∆x) - f(2) = (2 + ∆x)³ - 2(2 + ∆x) + 1 - (2³ - 2.2 + 1) = (∆x)³ + 6(∆x)² + 10∆x;

= (∆x)² + 6∆x + 10; = [(∆x)² + 6∆x + 10] = 10.

Vậy f'(2) = 10.

a)      

\(\begin{array}{l}f'({x_0}) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f({x_0})}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^2} - x_0^2}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^{2.\ln x}} - {e^{2.\ln {x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^{2.\ln {x_0}}}.\left( {{e^{2\ln x - 2\ln {x_0}}} - 1} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x_0^2\left( {{e^{2.\ln x - 2\ln {x_0}}} - 1} \right)}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x_0^2\left( {2\ln x - 2\ln {x_0}} \right)}}{{x - {x_0}}} = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \left( {\frac{x}{{{x_0}}}} \right)}}{{x - {x_0}}} = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \left( {1 + \frac{x}{{{x_0}}} - 1} \right)}}{{x - {x_0}}} = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{x}{{{x_0}}} - 1}}{{x - {x_0}}} = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{x - {x_0}}}{{{x_0}}}}}{{x - {x_0}}} = 2x_0^2\mathop {\lim }\limits_{x \to {x_0}} \frac{1}{{{x_0}}}\\ = 2x_0^2.\frac{1}{{{x_0}}} = 2x\\ \Rightarrow \left( {{x^2}} \right)' = 2x\end{array}\)

b) Dự đoán đạo hàm của hàm số \(y = {x^n}\) tại điểm x bất kì: \(y' = n.{x^{n - 1}}\)

\(\begin{array}{c}f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to  - 1} \frac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{ - {x^2} + 2x + 1 + 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{ - {x^2} + 2x + 3}}{{x + 1}}\\ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {3 - x} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \left( {3 - x} \right) = 3 + 1 = 4\end{array}\)

Vậy \(f'\left( { - 1} \right) = 4\)

a) Với bất kì \({x_0} \in \mathbb{R}\), ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x - {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} 1 = 1\)

Vậy \(f'\left( x \right) = {\left( x \right)^\prime } = 1\) trên \(\mathbb{R}\).

b) Ta có:

\(\begin{array}{l}{\left( {{x^2}} \right)^\prime } = 2{\rm{x}}\\{\left( {{x^3}} \right)^\prime } = 3{{\rm{x}}^2}\\...\\{\left( {{x^n}} \right)^\prime } = n{{\rm{x}}^{n - 1}}\end{array}\)