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Đề bài

Tìm đạo hàm cấp hai của mỗi hàm số sau:

a) $y=\frac{1}{2x+3}$.

b) $y={\log }_{3}x$.

c) $y=2^x$.

Lời giải chi tiết

a)

$y^{\prime }={\left(\frac{1}{2x-3}\right)}^{\prime }=\frac{1^{\prime }\left(2x+3\right)-1{\left(2x+3\right)}^{\prime }}{{\left(2x-3\right)}^2}=-\frac{2}{{\left(2x-3\right)}^2}$.

$y^{\prime \prime }={\left[-\frac{2}{{\left(2x-3\right)}^2}\right]}^{\prime }=-\frac{2^{\prime }{\left(2x-3\right)}^2-2{\left[{\left(2x-3\right)}^2\right]}^{\prime }}{{\left(2x-3\right)}^4}$

$=-\frac{-2.2{\left(2x-3\right)}^{\prime }\left(2x-3\right)}{{\left(2x-3\right)}^4}=\frac{8\left(2x-3\right)}{{\left(2x-3\right)}^4}=\frac{8}{{\left(2x-3\right)}^3}$.

b)

$y^{\prime }={\left({\log }_{3}x\right)}^{\prime }=\frac{1}{x\ln 3}$.

$y^{\prime \prime }={\left(\frac{1}{x\ln 3}\right)}^{\prime }=-\frac{{\left(x\ln 3\right)}^{\prime }}{{\left(x\ln 3\right)}^2}$

$=-\frac{\ln 3}{{\left(x\ln 3\right)}^2}=-\frac{\ln 3}{{\left(x\ln 3\right)}^2}=-\frac{1}{x.\ln 3}$.

c)

$y^{\prime }={\left(2^x\right)}^{\prime }=2^x.\ln 2$.

$y^{\prime \prime }={\left(2^x.\ln 2\right)}^{\prime }=2^x.\ln 2.\ln 2=2^x.{\left(\ln 2\right)}^2$.