Giải thích các bước giải:

5.Ta có:

$y'\\=(\sqrt{1+\cos x})' \\=\dfrac{1}{2\sqrt{1+\cos \left(x\right)}}\left(1+\cos \left(x\right)\right)'\:\\=\dfrac{1}{2\sqrt{1+\cos \left(x\right)}}\left(-\sin \left(x\right)\right)\\=-\dfrac{\sin \left(x\right)}{2\sqrt{1+\cos \left(x\right)}}$

6.Ta có:

$y'\\=(\dfrac{2x-3}{e^{4x}})'\\=\dfrac{\left(2x-3\right)'\:e^{4x}-\left(e^{4x}\right)'\:\left(2x-3\right)}{\left(e^{4x}\right)^2}\\=\dfrac{2e^{4x}-e^{4x}\cdot \:4\left(2x-3\right)}{\left(e^{4x}\right)^2}\\=\dfrac{2\left(-4x+7\right)}{e^{4x}}$

7.Ta có:

$y'\\=(\log_2(1+2^x))'\\=\dfrac{1}{\ln \left(2\right)}\left(\ln \left(1+2^x\right)\right)'\:\\=\dfrac{1}{1+2^x}\left(1+2^x\right)'\:\\=\dfrac{1}{\ln \left(2\right)}\cdot \:\dfrac{1}{1+2^x}\ln \left(2\right)\cdot \:2^x\\=\dfrac{2^x}{1+2^x}$

8.Ta có:

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

9.Ta có:

$y'\\=((x^2-x)e^x)'\\=\left(x^2-x\right)'\:e^x+\left(e^x\right)'\:\left(x^2-x\right)\\=\left(2x-1\right)e^x+e^x\left(x^2-x\right)\\=e^xx^2+e^xx-e^x$

10.Ta có:

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