mọi người giúp em vs ạ , Toán Lớp12

Giải thích các bước giải:

a.Ta có:

$y'=(\log_2(9x-5))'$

$=\dfrac{1}{\ln \left(2\right)}\left(\ln \left(9x-5\right)\right)'$

$=\dfrac{1}{\ln \left(2\right)}\left(\ln \left(9x-5\right)\right)'\:$

$=\dfrac{9}{\ln \left(2\right)\left(9x-5\right)}$

$\to a$ đúng

b.Ta có:

$y'=(2e^{3x+1})'=2\left(e^{3x+1}\right)'\:=2\cdot e^{3x+1}\left(3x+1\right)'\:=2\cdot e^{3x+1}\cdot 3=6e^{3x+1} $

$\to b$ đúng

c.Ta có:

$y'=(3^{x^3-1})'$

$\to y'=\left(e^{\left(x^3-1\right)\ln \left(3\right)}\right)'\:$

$\to y'=e^{\left(x^3-1\right)\ln \left(3\right)}\left(\left(x^3-1\right)\ln \left(3\right)\right)'\:$

$\to y'=e^{\left(x^3-1\right)\ln \left(3\right)}\cdot \:3\ln \left(3\right)x^2$

$\to y'=3\ln \left(3\right)\cdot \:3^{x^3-1}x^2$

$\to c$ đúng

d.Ta có:

$y'=(\ln\sqrt{x})'=\dfrac{1}{\sqrt{x}}\left(\sqrt{x}\right)'\:=\dfrac{1}{\sqrt{x}}\cdot \dfrac{1}{2\sqrt{x}}=\dfrac1{2x}$

$\to d$ sai