tính đạo hàm sau
\(y=ln\left(\frac{1-x^2}{1+x^2}\right)\)
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14 tháng 8 2023
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: \(y=ln^2x\)
=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
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 \(\left( {{{\log }_a}x} \right)' = \left( {\frac{{\ln x}}{{\ln a}}} \right)' = \frac{1}{{x\ln a}}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
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 \(\left( {0; + \infty } \right)\).
DT
1
12 tháng 5 2016
\(y'=\frac{2x}{x^2+1}+\frac{2x-1}{\left(x^2-x+1\right)\ln2}\)
ML
1
VB
1
5 tháng 5 2016
xét hàm số y=ln(\(x+\sqrt{1+x^2}\))
Ta có
y'=\(\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)=\frac{1}{x+\sqrt{1+x^2}}.\frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 11 2021
a.
\(y'=\dfrac{\left(1+\sqrt{3x-1}\right)'}{1+\sqrt{3x-1}}=\dfrac{3}{2\left(1+\sqrt{3x-1}\right)\sqrt{3x-1}}\)
b.
\(y'=\dfrac{\left(2sin^2x-1\right)'}{\left(2sin^2x-1\right).ln10}=\dfrac{2sin2x}{\left(2sin^2x-1\right)ln10}\)
c.
\(y'=\left(3x^2+3\right)3^{x^3+3x+1}.e^x.ln3+3^{x^3+3x+1}.e^x\)
HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023
\(a,y'=\left(\dfrac{\sqrt{x}}{x+1}\right)'\\ =\dfrac{\left(\sqrt{x}\right)'\left(x+1\right)-\sqrt{x}\left(x+1\right)}{\left(x+1\right)^2}\\ =\dfrac{\dfrac{x+1}{2\sqrt{x}}-\sqrt{x}}{\left(x+1\right)^2}\\ =\dfrac{x+1-2x}{2\sqrt{x}\left(x+1\right)^2}\\ =\dfrac{-x+1}{2\sqrt{x}\left(x+1\right)^2}\)
\(b,y'=\left(\sqrt{x}+1\right)'\left(x^2+2\right)+\left(\sqrt{x}+1\right)\left(x^2+2\right)'\\ =\dfrac{x^2+2}{2\sqrt{x}}+\left(\sqrt{x}+1\right)\cdot2x\)
2
18 tháng 11 2023
`a)TXĐ:R\\{1;1/3}`
`y'=[-4(6x-4)]/[(3x^2-4x+1)^5]`
`b)TXĐ:R`
`y'=2x. 3^[x^2-1] ln 3-e^[-x+1]`
`c)TXĐ: (4;+oo)`
`y'=[2x-4]/[x^2-4x]+2/[(2x-1).ln 3]`
`d)TXĐ:(0;+oo)`
`y'=ln x+2/[(x+1)^2].2^[[x-1]/[x+1]].ln 2`
`e)TXĐ:(-oo;-1)uu(1;+oo)`
`y'=-7x^[-8]-[2x]/[x^2-1]`
AH
Akai Haruma
Giáo viên
18 tháng 11 2023
Lời giải:
a.
$y'=-4(3x^2-4x+1)^{-5}(3x^2-4x+1)'$
$=-4(3x^2-4x+1)^{-5}(6x-4)$
$=-8(3x-2)(3x^2-4x+1)^{-5}$
b.
$y'=(3^{x^2-1})'+(e^{-x+1})'$
$=(x^2-1)'3^{x^2-1}\ln 3 + (-x+1)'e^{-x+1}$
$=2x.3^{x^2-1}.\ln 3 -e^{-x+1}$
c.
$y'=\frac{(x^2-4x)'}{x^2-4x}+\frac{(2x-1)'}{(2x-1)\ln 3}$
$=\frac{2x-4}{x^2-4x}+\frac{2}{(2x-1)\ln 3}$
d.
\(y'=(x\ln x)'+(2^{\frac{x-1}{x+1}})'=x(\ln x)'+x'\ln x+(\frac{x-1}{x+1})'.2^{\frac{x-1}{x+1}}\ln 2\)
\(=x.\frac{1}{x}+\ln x+\frac{2}{(x+1)^2}.2^{\frac{x-1}{x+1}}\ln 2\\ =1+\ln x+\frac{2^{\frac{2x}{x+1}}\ln 2}{(x+1)^2}\)
e.
\(y'=-7x^{-8}-\frac{(x^2-1)'}{x^2-1}=-7x^{-8}-\frac{2x}{x^2-1}\)
ta có:
\(y'=\frac{\left(\frac{1-x^2}{1+x^2}\right)'}{\frac{1-x^2}{1+x^2}}=\frac{\frac{-2x.\left(1+x^2\right)-2x.\left(1-x^2\right)}{\left(1+x^2\right)^2}}{\frac{1-x^2}{1+x^2}}=\frac{\frac{-4x}{\left(1+x^2\right)^2}}{\frac{1-x^2}{1+x^2}}=\frac{-4x}{\left(1+x^2\right)\left(1-x^2\right)}=\frac{-4x}{1-x^4}\)