y '   =   1 x 1   -   x 2

y '   =   - 1 x - 2 2

y' = 4 - 2x

y '   =   3 2 3 x + 1

y' = 3

a) Với \({x_0}\) bất kì, ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{k{x^2} + c - \left( {kx_0^2 + c} \right)}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{k\left( {{x^2} - x_0^2} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{k\left( {x - {x_0}} \right)\left( {x + {x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \left[ {k\left( {x + {x_0}} \right)} \right] = 2k{x_0}\)

Vậy hàm số \(y = k{x^2} + c\) có đạo hàm là hàm số \(y' = 2kx\)

b) Với \({x_0}\) bất kì, ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^3} - x_0^3}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x - {x_0}} \right)\left( {{x^2} + x{x_0} + x_0^2} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \left( {{x^2} + x{x_0} + x_0^2} \right) = 3x_0^2\)

Vậy hàm số \(y = {x^3}\) có đạo hàm là hàm số \(y' = 3{x^2}\)

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

\(=\lim\limits_{x\rightarrow x0}\dfrac{sinx-sin\left(x0\right)}{x-x0}\)

\(=\lim\limits_{x\rightarrow x0}\dfrac{2\cdot cos\left(\dfrac{x+x0}{2}\right)\cdot sin\left(\dfrac{x-x0}{2}\right)}{x-x_0}\)

\(=\lim\limits_{x\rightarrow x0}\dfrac{2\cdot sin\left(\dfrac{x-x_0}{2}\right)\cdot cos\left(\dfrac{x+x_0}{2}\right)}{x-x_0}\)

\(=\lim\limits_{x\rightarrow x0}\dfrac{cos\left(x+x_0\right)}{2}=cos\left(x0\right)\)

=>\(\left(sinx'\right)=cosx\)

\(\begin{array}{l}f'(x) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x + {x_0}) - f(x)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{x + {x_0}}} - {e^x}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{x + {x_0}}} - {e^x}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}({e^{{x_0}}} - 1)}}{x} = {e^x}.\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x_0}}} - 1}}{x} = {e^x}.1 = {e^x}\\ \Rightarrow f'(x) = {e^x}\end{array}\)

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln x - \ln {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \frac{x}{{{x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{\ln \frac{x}{{{x_0}}}}}{{\ln e}}}}{{x - {x_0}}} = \frac{1}{{\ln e}}.\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \frac{x}{{{x_0}}}}}{{x - {x_0}}}\\ = \frac{1}{{\ln e}}\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \left( {1 + \frac{x}{{{x_0}}} - 1} \right)}}{{x - {x_0}}} = \frac{1}{{\ln e}}\mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{x}{{{x_0}}} - 1}}{{x - {x_0}}} = \frac{1}{{\ln e}}.\mathop {\lim }\limits_{u \to 0} \frac{{\frac{{x - {x_0}}}{{{x_0}}}}}{{x - {x_0}}} = \frac{1}{{{x_0}\ln e}}\\ \Rightarrow \left( {\ln x} \right)' = \frac{1}{{x\ln e}} = \frac{1}{x}\end{array}\)