Advertisements (Quảng cáo)
Tính đạo hàm của mỗi hàm số sau :
a. \(y = {1 \over {2x – 1}}\,\text{ với }\,x \ne {1 \over 2}\)
b. \(y = \sqrt {3 – x} \) với \(x < 3\).
a. Đặt \(f(x)=y = {1 \over {2x – 1}}\)
Với \({x_0} \ne {1 \over 2}\) ta có:
Advertisements (Quảng cáo)
\(\eqalign{ & f’\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} = {{f\left( {{x_0} + \Delta x} \right) – f\left( {{x_0}} \right)} \over {\Delta x}} \cr & = \mathop {\lim }\limits_{\Delta x \to 0} {{{1 \over {2{x_0} + 2\Delta x – 1}} – {1 \over {2{x_0} – 1}}} \over {\Delta x}} \cr & = \mathop {\lim }\limits_{\Delta x \to 0} {{ – 2\Delta x} \over {\Delta x\left( {2{x_0} + 2\Delta x – 1} \right)\left( {2{x_0} – 1} \right)}} \cr & = \mathop {\lim }\limits_{\Delta x \to 0} {{ – 2} \over {\left( {2{x_0} + 2\Delta x – 1} \right)\left( {2{x_0} – 1} \right)}} \cr & = {{ – 2} \over {{{\left( {2{x_0} – 1} \right)}^2}}} \cr} \)
b. Đặt \(f(x)=y = \sqrt {3 – x} \)
Với x0 < 3, ta có:
\(\eqalign{ & f’\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} {{f\left( {{x_0} + \Delta x} \right) – f\left( {{x_0}} \right)} \over {\Delta x}} \cr & = \mathop {\lim }\limits_{\Delta x \to 0} {{\sqrt {3 – {x_0} – \Delta x} – \sqrt {3 – {x_0}} } \over {\Delta x}} \cr & = \mathop {\lim }\limits_{\Delta x \to 0} {{ – 1} \over {\sqrt {3 – {x_0} – \Delta x} + \sqrt {3 – {x_0}} }} = {{ – 1} \over {2\sqrt {3 – {x_0}} }} \cr} \)