\(f'\left(x\right)=\dfrac{1}{x\cdot ln10}\)

=>\(f'\left(\dfrac{1}{2}\right)=\dfrac{1}{\dfrac{1}{2}\cdot ln10}=\dfrac{2}{ln10}\)

a: \(y'=\left(x^2+3x-1\right)'\cdot e^x+\left(x^2+3x-1\right)\cdot\left(e^x\right)'\)

\(=e^x\left(2x+3\right)+\left(x^2+3x-1\right)\cdot e^x\)

\(=e^x\left(x^2+5x+2\right)\)

b: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)

\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)

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 \(y' = 3{{\rm{e}}^{3{\rm{x}} + 1}}\).

tham khảo:

a)y′=2\(^{3x-x^2}\).ln2.(3−2x)

b) y′\(\dfrac{4}{ln3}\).\(\dfrac{1}{4x+1}\).4=\(\dfrac{4}{\left(4x+1\right)ln3}\)

a, \(y=\left(2x^3+3\right)^2\)

\(y'=2\left(2x^3+3\right)6x^2\)

\(=12x^2\left(2x^3+3\right)\)

b,\(y=cos3x\)

\(y'=-3sin3x\)

c, \(y=log_2\left(x^2+2\right)\)

\(y'=\dfrac{2x}{\left(x^2+2\right)ln2}\)

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.

\(y'=\left(3x^2+3\right)3^{x^3+3x+1}.e^x.ln3+3^{x^3+3x+1}.e^x\)

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)\(y'\left(x\right)=\dfrac{1}{x\sqrt{x}}.\left(+\dfrac{1}{2\sqrt{x}}\right)\)

\(=\dfrac{1}{\sqrt{x}\left(2\sqrt{x}+\sqrt{x}+2\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(3\sqrt{x}+2\right)}\)

\(y=\log\left(\frac{1-\sqrt{x}}{2\sqrt{x}}\right)\Rightarrow y'=\frac{\left(\frac{1-\sqrt{x}}{2\sqrt{x}}\right)'}{\frac{1-\sqrt{x}}{x^2}\ln10}=\frac{-\frac{1}{2\sqrt{x}}.2\sqrt{x}-\frac{1}{\sqrt{x}}.\left(1-\sqrt{x}\right)}{\frac{1-\sqrt{x}}{2\sqrt{x}}\ln10}\)

                                 \(=\frac{-1-\frac{1-\sqrt{x}}{\sqrt{x}}}{4x.\frac{1-\sqrt{x}}{2\sqrt{x}}\ln10}=\frac{1}{2x\left(\sqrt{x}-1\right)\ln10}\)

\(y'=\left(log_2\left(2x-1\right)\right)'=\dfrac{2}{\left(2x-1\right)ln2}\)