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tham khảo:
a)\(y'=xsin2x+sin^2x\)
\(y'=sin^2x+xsin2x\)
b)\(y'=-2sin2x+2cosx\\ y'=2\left(cosx-sin2x\right)\)
c)\(y=sin3x-3sinx\)
\(y'=3cos3x-3cosx\)
d)\(y'=\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\)
\(y'=\dfrac{sin^2x-cos^2x}{sin^2x.cos^2x}\)
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: \(y'=\left(sin3x\right)'+\left(sin^2x\right)'=3\cdot cos3x+sin\left(x+pi\right)\)
b: \(y'=\left(log_2\left(2x+1\right)\right)'+\left(3^{-2x+1}\right)'\)
\(=\dfrac{2}{\left(2n+1\right)\cdot ln2}-2\cdot3^{-2x+1}\cdot ln3\)
\(a,y'=\left(f\left(g\left(x\right)\right)\right)'\)
\(=f'\left(g\left(x\right)\right).g'\left(x\right)\)
\(=e^{g\left(x\right)}.\left(2x-1\right)\)
\(=e^{x^2-x}.\left(2x-1\right)\)
\(b,y'=\dfrac{d}{dx}\left(3^{sinx}\right)\)
\(=\dfrac{d}{dx}\left(e^{ln3.sinx}\right)\)
\(=\dfrac{d}{dx}\left(ln3.sinx\right).e^{ln3.sinx}\)
\(=ln3.cosx.3^{sinx}\)
tham khảo:
a)\(y'\left(x\right)=5\left(\dfrac{2x-1}{x+2}\right)^4.\dfrac{\left(x+2\right)\left(2\right)-\left(2x-1\right).1}{\left(x+2\right)^2}\)
\(=\dfrac{10\left(2x-1\right)\left(x+2\right)^3}{\left(x+2\right)^4}=\dfrac{20x-50}{\left(x+2\right)^4}\)
b)\(y'\left(x\right)=\dfrac{2\left(x^2+1\right)-2x\left(2x\right)}{\left(x^2+1\right)^2}\)\(=\dfrac{2\left(1-x^2\right)}{\left(x^2+1\right)^2}\)
c)\(y'\left(x\right)=e^x.2sinxcosx+e^xsin^2x.2cosx\)
\(=2e^xsinx\left(cosx+sinxcosx\right)\)
\(=2e^xsinxcos^2x\)
d)\(y'\left(x\right)=\dfrac{1}{x\sqrt{x}}.\left(+\dfrac{1}{2\sqrt{x}}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(2\sqrt{x}+\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(3\sqrt{x}+2\right)}\)
\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
\(f'\left(x0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x_0\right)}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{sinx-sin\left(x0\right)}{x-x0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{2\cdot cos\left(\dfrac{x+x0}{2}\right)\cdot sin\left(\dfrac{x-x0}{2}\right)}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{2\cdot sin\left(\dfrac{x-x_0}{2}\right)\cdot cos\left(\dfrac{x+x_0}{2}\right)}{x-x_0}\)
\(=\lim\limits_{x\rightarrow x0}\dfrac{cos\left(x+x_0\right)}{2}=cos\left(x0\right)\)
=>\(\left(sinx'\right)=cosx\)
- Bước đầu tiên ta áp dụng công thức ( u α ) ' với
- Ta có :
- Tính :
Chọn D.