tham khảo:

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

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

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

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

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 \(f'\left( { - 1} \right) = 3\)

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: \(y' = n.{x^{n - 1}}\)

\(\begin{array}{l}\Delta x = x - {x_0} = x - 1\\\Delta y = f({x_0} + \Delta x) - f({x_0}) = f(x) - f(1)\\\mathop {\lim }\limits_{x \to 1} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{x \to 1} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^3} - 1 - (3 - 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^3} - 3}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{3(x - 1)({x^2} + x + 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} (3({x^2} + x + 1)) = 9\end{array}\)

Vậy \(f'(1) = 9\)

\(f'\left(3\right)=\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-f\left(3\right)}{x-3}\\ =\lim\limits_{x\rightarrow3}\dfrac{2x-6}{x-3}\\ =2\)

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)    Ta có: \(f'\left( x \right) = \left( {{x^{22}}} \right)' = 22.{x^{21}}\)

b)    Đạo hàm của hàm số tại điểm \({x_0} =  - 1\) là: \(f'\left( { - 1} \right) = 22.{\left( { - 1} \right)^{21}} =  - 22\)

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:

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

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