Bài 4 trang 76

Đề bài

Tính đạo hàm cấp hai của mỗi hàm số sau:

a) $y=2x^4-3x^3+5x^2$;

b) $y=\frac{2}{3-x}$;

c) $y=\sin 2x\cos x$;

d) $y=e^{-2x+3}$;

e) $y=\ln (x+1)$;

f) $y=\ln (e^x+1)$.

Lời giải chi tiết

a) $y=2x^4-3x^3+5x^2$

$\Rightarrow y^{\prime }=8x^3-9x^2+10x$

$\Rightarrow y^{\prime \prime }=24x^2-18x+10$.

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

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

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

c) $y=\sin 2x\cos x$

$\Rightarrow y^{\prime }=2\cos 2x.\cos x-\sin 2x.\sin x$

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

$=\frac{3}{2}\cos 3x+\frac{1}{2}\cos x$

$\Rightarrow y^{\prime \prime }=-\frac{3}{2}.3.\sin 3x-\frac{1}{2}\sin x$

$=-\frac{9}{2}\sin 3x-\frac{1}{2}\sin x$.

d) $y=e^{-2x+3}\Rightarrow y^{\prime }=-2e^{-2x+3}$

$\Rightarrow y^{\prime \prime }=4.e^{-2x+3}$.

e) $y=\ln (x+1)\Rightarrow y^{\prime }=\frac{1}{x+1}$

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

f) $y=\ln (e^x+1)\Rightarrow y^{\prime }=\frac{e^x}{e^x+1}$

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

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