Đáp án: a.$ y'=\dfrac12.\dfrac{1}{\cos^2\dfrac{x}{2}}-\dfrac12.\dfrac{1}{\sin^2\dfrac{x}{2}}$
b.$y'=\dfrac{-x+3}{2\left(1-x\right)\sqrt{1-x}}$
Giải thích các bước giải:
a.Ta có :
$y=\tan\dfrac{x}{2}+\cot\dfrac{x}{2}$
$\to y'=(\tan\dfrac{x}{2}+\cot\dfrac{x}{2})'$
$\to y'=\dfrac12.\dfrac{1}{\cos^2\dfrac{x}{2}}-\dfrac12.\dfrac{1}{\sin^2\dfrac{x}{2}}$
b.Ta có :
$y=\dfrac{x+1}{\sqrt{1-x}}$
$\to y'=(\dfrac{x+1}{\sqrt{1-x}})'$
$\to y'=\dfrac{\left(x+1\right)'\sqrt{1-x}-\left(\sqrt{1-x}\right)'\left(x+1\right)}{\left(\sqrt{1-x}\right)^2}$
$\to y'=\dfrac{1\cdot \sqrt{1-x}-\left(-\dfrac{1}{2\sqrt{1-x}}\right)\left(x+1\right)}{\left(\sqrt{1-x}\right)^2}$
$\to y'=\dfrac{-x+3}{2\left(1-x\right)\sqrt{1-x}}$
