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Đặt \(g\left(x\right)=\left(1+x\right)\left(2+x\right)...\left(2017+x\right)\)
\(\Rightarrow g\left(0\right)=1.2.3...2017=2017!\)
\(f\left(x\right)=\dfrac{x}{g\left(x\right)}\Rightarrow f'\left(x\right)=\dfrac{g\left(x\right)-x.g'\left(x\right)}{g^2\left(x\right)}\)
\(\Rightarrow f'\left(0\right)=\dfrac{g\left(0\right)-0.g'\left(x\right)}{\left[g\left(0\right)\right]^2}=\dfrac{g\left(0\right)}{\left[g\left(0\right)\right]^2}=\dfrac{1}{g\left(0\right)}=\dfrac{1}{2017!}\)
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) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)
\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)
\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)
\(\begin{array}{l}f'\left( x \right) = {\left( {\sqrt x } \right)'} = \frac{1}{{2\sqrt x }}\\ \Rightarrow f'\left( 9 \right) = \frac{1}{{2\sqrt 9 }} = \frac{1}{{2.3}} = \frac{1}{6}\end{array}\)
\(f'\left( x \right) = {10^x}.\ln 10 \Rightarrow f'\left( { - 1} \right) = {10^{ - 1}}.\ln 10 = \frac{{\ln 10}}{{10}}\)
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 \(h'\left( {{x_0}} \right) = f'\left( {{x_0}} \right) + g'\left( {{x_0}} \right)\).
\(f'\left( x \right) = \frac{1}{{{{\cos }^2}x}} \Rightarrow f'\left( { - \frac{\pi }{6}} \right) = \frac{1}{{{{\cos }^2}\left( { - \frac{\pi }{6}} \right)}} = \frac{4}{3}\)
\(f'\left( x \right) = - \frac{1}{{{{\sin }^2}x}} \Rightarrow f'\left( { - \frac{\pi }{3}} \right) = - \frac{1}{{{{\sin }^2}\left( { - \frac{\pi }{3}} \right)}} = - \frac{4}{3}\)
\(y = \left| x \right| = \left\{ \begin{array}{l}x\,\,\,(x \ge 0)\\ - x\,\,\,(x < 0)\end{array} \right. \Rightarrow y' = \left\{ \begin{array}{l}1\,\,\,(x \ge 0)\\ - 1\,\,\,(x < 0)\end{array} \right.\)
Ta có: \(\mathop {\lim }\limits_{x \to {0^ + }} y' = 1 \ne - 1 = \mathop {\lim }\limits_{x \to {0^ - }} y'\)
Vậy không tồn tại đạo hàm của hàm số tại x = 0