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\(f'\left(x0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x_0\right)}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{sinx-sin\left(x0\right)}{x-x0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{2\cdot cos\left(\dfrac{x+x0}{2}\right)\cdot sin\left(\dfrac{x-x0}{2}\right)}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{2\cdot sin\left(\dfrac{x-x_0}{2}\right)\cdot cos\left(\dfrac{x+x_0}{2}\right)}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{cos\left(x+x_0\right)}{2}=cos\left(x0\right)\)
=>\(\left(sinx'\right)=cosx\)
a) Với x bất kì và \(h = x - {x_0}\), ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{{x_0} + h}} - {e^{{x_0}}}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{{x_o}}}\left( {{e^h} - 1} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} {e^{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h} = {e^{{x_0}}}\end{array}\)
Vậy hàm số \(y = {e^x}\) có đạo hàm là hàm số \(y' = {e^x}\)
b) Ta có \({a^x} = {e^{x\ln a}}\,\)nên \(\left( {{a^x}} \right)' = \left( {{e^{x\ln a}}} \right)' = \left( {x\ln a} \right)'.{e^{x\ln a}} = {e^{x\ln a}}\ln a = {a^x}\ln a\)
Với bất kì \({x_0} \in \mathbb{R}\), ta có:
\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin x - \sin {x_0}}}{{x - {x_0}}}\)
Đặt \(x = {x_0} + \Delta x\). Ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left( {{x_0} + \Delta x} \right) - \sin {x_0}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\cos \Delta x + \cos {x_0}\sin \Delta x - \sin {x_0}}}{{\Delta x}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\cos \Delta x - \sin {x_0}}}{{\Delta x}} + \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos {x_0}\sin \Delta x}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {\cos \Delta x - 1} \right)}}{{\Delta x}} + \mathop {\lim }\limits_{\Delta x \to 0} \cos {x_0}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}}\end{array}\)
Lại có:
\(\begin{array}{l}\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {\cos \Delta x - 1} \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {\cos \Delta x - 1} \right)\left( {\cos \Delta x + 1} \right)}}{{\Delta x\left( {\cos \Delta x + 1} \right)}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( {{{\cos }^2}\Delta x - 1} \right)}}{{\Delta x\left( {\cos \Delta x + 1} \right)}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}\left( { - {{\sin }^2}\Delta x} \right)}}{{\Delta x\left( {\cos \Delta x + 1} \right)}} = - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin {x_0}.\sin \Delta x}}{{\left( {\cos \Delta x + 1} \right)}} = - 1.\frac{{\sin {x_0}.\sin 0}}{{\cos 0 + 1}} = 0\\\mathop {\lim }\limits_{\Delta x \to 0} \cos {x_0}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \Delta x}}{{\Delta x}} = \cos {x_0}.1 = \cos {x_0}\end{array}\)
Vậy \(f'\left( {{x_0}} \right) = \cos {x_0}\)
Vậy \(f'\left( x \right) = \cos x\) trên \(\mathbb{R}\).
a) Với x > 0 bất kì và \(h = x - {x_0}\) ta có
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{x_0} + h} \right) - \ln {x_0}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}.{x_0}}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}}} = \frac{1}{{{x_0}}}\end{array}\)
Vậy hàm số \(y = \ln x\) có đạo hàm là hàm số \(y' = \frac{1}{x}\)
b) Ta có \({\log _a}x = \frac{{\ln x}}{{\ln a}}\) nên \(\left( {{{\log }_a}x} \right)' = \left( {\frac{{\ln x}}{{\ln a}}} \right)' = \frac{1}{{x\ln a}}\)
\(\begin{array}{l}f'(x) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x + {x_0}) - f(x)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{x + {x_0}}} - {e^x}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{x + {x_0}}} - {e^x}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}({e^{{x_0}}} - 1)}}{x} = {e^x}.\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x_0}}} - 1}}{x} = {e^x}.1 = {e^x}\\ \Rightarrow f'(x) = {e^x}\end{array}\)
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f(x) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln x - \ln {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \frac{x}{{{x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{\ln \frac{x}{{{x_0}}}}}{{\ln e}}}}{{x - {x_0}}} = \frac{1}{{\ln e}}.\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \frac{x}{{{x_0}}}}}{{x - {x_0}}}\\ = \frac{1}{{\ln e}}\mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln \left( {1 + \frac{x}{{{x_0}}} - 1} \right)}}{{x - {x_0}}} = \frac{1}{{\ln e}}\mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{x}{{{x_0}}} - 1}}{{x - {x_0}}} = \frac{1}{{\ln e}}.\mathop {\lim }\limits_{u \to 0} \frac{{\frac{{x - {x_0}}}{{{x_0}}}}}{{x - {x_0}}} = \frac{1}{{{x_0}\ln e}}\\ \Rightarrow \left( {\ln x} \right)' = \frac{1}{{x\ln e}} = \frac{1}{x}\end{array}\)
a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó
\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)
b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)
c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)
a) Với bất kì \({x_0} \in \mathbb{R}\), ta có:
\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{e^x} - {e^{{x_0}}}}}{{x - {x_0}}}\)
Đặt \(x = {x_0} + \Delta x\). Ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0} + \Delta x}} - {e^{{x_0}}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0}}}.{e^{\Delta x}} - {e^{{x_0}}}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{{x_0}}}.\left( {{e^{\Delta x}} - 1} \right)}}{{\Delta x}}\\ & = \mathop {\lim }\limits_{\Delta x \to 0} {e^{{x_0}}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{\Delta x}} - 1}}{{\Delta x}} = {e^{{x_0}}}.1 = {e^{{x_0}}}\end{array}\)
Vậy \({\left( {{e^x}} \right)^\prime } = {e^x}\) trên \(\mathbb{R}\).
b) Với bất kì \({x_0} > 0\), ta có:
\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\ln {\rm{x}} - \ln {{\rm{x}}_0}}}{{x - {x_0}}}\)
Đặt \(x = {x_0} + \Delta x\). Ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{x_0} + \Delta x} \right) - \ln {{\rm{x}}_0}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {\frac{{{x_0} + \Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\Delta x}}\\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}}.\frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}}\end{array}\)
Đặt \(\frac{{\Delta x}}{{{x_0}}} = t\). Lại có: \(\mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{{x_0}}} = \frac{1}{{{x_0}}};\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{{{{\rm{x}}_0}}}} \right)}}{{\frac{{\Delta x}}{{{x_0}}}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\ln \left( {1 + t} \right)}}{t} = 1\)
Vậy \(f'\left( {{x_0}} \right) = \frac{1}{{{x_0}}}.1 = \frac{1}{{{x_0}}}\)
Vậy \({\left( {\ln x} \right)^\prime } = \frac{1}{x}\) trên khoảng \(\left( {0; + \infty } \right)\).
a) Tập xác định của hàm số là \(D = \mathbb{R}\)
Do đó, nếu x thuộc tập xác định D thì –x cũng thuộc tập xác định D
Ta có: \(f\left( { - x} \right) = \sin \left( { - x} \right) = - \sin x = - f\left( x \right),\;\forall x\; \in \;D\)
Vậy \(y = \sin x\) là hàm số lẻ.
b)
\(x\) | \( - \pi \) | \( - \frac{{3\pi }}{4}\) | \( - \frac{\pi }{2}\) | \( - \frac{\pi }{4}\) | 0 | \(\frac{\pi }{4}\) | \(\frac{\pi }{2}\) | \(\frac{{3\pi }}{4}\) | \(\pi \) |
\(\sin x\) | \(0\) | \( - \frac{{\sqrt 2 }}{2}\) | \( - 1\) | \( - \frac{{\sqrt 2 }}{2}\) | 0 | \(\frac{{\sqrt 2 }}{2}\) | 1 | \(\frac{{\sqrt 2 }}{2}\) | 0 |
c) Từ đồ thị trên, ta thấy hàm số \(y = \sin x\) có tập xác định là \(\mathbb{R}\), tập giá trị là [-1;1] và đồng biến trên mỗi khoảng \(\left( { - \frac{\pi }{2} + k2\pi ;\frac{\pi }{2} + k2\pi } \right)\) và nghịch biến trên mỗi khoảng \(\left( {\frac{\pi }{2} + k2\pi ;\frac{{3\pi }}{2} + k2\pi } \right),\;k\; \in \;\mathbb{Z}.\)
a) \(\sin \left( {x + h} \right) - \sin x = 2\cos \frac{{2x + h}}{2}.\sin \frac{h}{2}\)
b) Với \({x_0}\) bất kì, ta có:
\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin x - \sin {x_0}}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{2\cos \frac{{x + {x_0}}}{2}.\sin \frac{{x - {x_0}}}{2}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sin \frac{{x - {x_0}}}{2}}}{{\frac{{x - {x_0}}}{2}}}.\mathop {\lim }\limits_{x \to {x_0}} \cos \frac{{x + {x_0}}}{2} = \cos {x_0}\end{array}\)
Vậy hàm số y = sin x có đạo hàm là hàm số \(y' = \cos x\)