Đạo hàm của hàm số y = e x - e - x là:
A. y ' = e x + e - x 2 x
B. y ' = e x - e - x 2 x
C. y ' = e x - e - x x
D. y ' = e x + e - x x
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\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
\(a,y'=\left(f\left(g\left(x\right)\right)\right)'\)
\(=f'\left(g\left(x\right)\right).g'\left(x\right)\)
\(=e^{g\left(x\right)}.\left(2x-1\right)\)
\(=e^{x^2-x}.\left(2x-1\right)\)
\(b,y'=\dfrac{d}{dx}\left(3^{sinx}\right)\)
\(=\dfrac{d}{dx}\left(e^{ln3.sinx}\right)\)
\(=\dfrac{d}{dx}\left(ln3.sinx\right).e^{ln3.sinx}\)
\(=ln3.cosx.3^{sinx}\)
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 \(\left( {{a^x}} \right)' = \left( {{e^{x\ln a}}} \right)' = \left( {x\ln a} \right)'.{e^{x\ln a}} = {e^{x\ln a}}\ln a = {a^x}\ln a\)
a: \(y'=\left(x^2+2x\right)'\left(x^3-3x\right)+\left(x^2+2x\right)\left(x^3-3x\right)'\)
\(=\left(2x+2\right)\left(x^3-3x\right)+\left(x^2+2x\right)\left(3x^2-3\right)\)
\(=2x^4-6x^2+2x^3-6x+3x^4-3x^2+6x^3-6x\)
\(=5x^4+8x^3-9x^2-12x\)
b: y=1/-2x+5
=>\(y'=\dfrac{2}{\left(2x+5\right)^2}\)
c: \(y'=\dfrac{\left(4x+5\right)'}{2\sqrt{4x+5}}=\dfrac{4}{2\sqrt{4x+5}}=\dfrac{2}{\sqrt{4x+5}}\)
d: \(y'=\left(sinx\right)'\cdot cosx+\left(sinx\right)\cdot\left(cosx\right)'\)
\(=cos^2x-sin^2x=cos2x\)
e: \(y=x\cdot e^x\)
=>\(y'=e^x+x\cdot e^x\)
f: \(y=ln^2x\)
=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)
\(a,y'=8x^3-10x\\ \Rightarrow y''=24x^2-10\\ b,y'=e^x+xe^x\\ \Rightarrow y''=e^x+e^x+xe^x=2e^x+xe^x\)
a: \(y'=4\cdot3x^2-3\cdot2x+2=12x^2-6x+2\)
b: \(y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}=\dfrac{x-1-x-1}{\left(x-1\right)^2}=\dfrac{-2}{\left(x-1\right)^2}\)
c: \(y'=-2\cdot\left(\sqrt{x}\cdot x\right)'\)
\(=-2\cdot\left(\dfrac{x+x}{2\sqrt{x}}\right)=-2\cdot\dfrac{2x}{2\sqrt{x}}=-2\sqrt{x}\)
d: \(y'=\left(3sinx+4cosx-tanx\right)\)'
\(=3cosx-4sinx+\dfrac{1}{cos^2x}\)
e: \(y'=\left(4^x+2e^x\right)'\)
\(=4^x\cdot ln4+2\cdot e^x\)
f: \(y'=\left(x\cdot lnx\right)'=lnx+1\)
\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)
a) \(y' = {\left( {{x^2} - x} \right)^\prime }{.2^x} + \left( {{x^2} - x} \right).{\left( {{2^x}} \right)^\prime } = \left( {2{\rm{x}} - 1} \right){.2^x} + \left( {{x^2} - x} \right){.2^x}.\ln 2\).
b) \(y' = {\left( {{x^2}} \right)^\prime }.{\log _3}x + {x^2}.{\left( {{{\log }_3}x} \right)^\prime } = 2{\rm{x}}.{\log _3}x + {x^2}.\frac{1}{{x\ln 3}} = 2{\rm{x}}.{\log _3}x + \frac{x}{{\ln 3}}\).
c) Đặt \(u = 3{\rm{x}} + 1\) thì \(y = {e^u}\). Ta có: \(u{'_x} = {\left( {3{\rm{x}} + 1} \right)^\prime } = 3\) và \(y{'_u} = {\left( {{e^u}} \right)^\prime } = {e^u}\).
Suy ra \(y{'_x} = y{'_u}.u{'_x} = {e^u}.3 = 3{{\rm{e}}^{3{\rm{x}} + 1}}\).
Vậy \(y' = 3{{\rm{e}}^{3{\rm{x}} + 1}}\).
Chọn A.
Ta có:![](http://cdn.hoc24.vn/bk/5mKLUbzcHv5W.png)
![](http://cdn.hoc24.vn/bk/KHPQOGVF5ldM.png)