Giải thích các bước giải:
a.Ta có:
$y=x\sqrt{x^2+1}$
$\to y'=(x\sqrt{x^2+1})'$
$\to y'=x'\sqrt{x^2+1}+\left(\sqrt{x^2+1}\right)'\:x$
$\to y'=1\cdot \sqrt{x^2+1}+\dfrac{x}{\sqrt{x^2+1}}\cdot x$
$\to y'=\dfrac{2x^2+1}{\sqrt{x^2+1}}$
b.Ta có:
$y=\dfrac{3}{(2x+5)^2}$
$\to y'=(\dfrac{3}{(2x+5)^2})'$
$\to y'=3\left(\left(2x+5\right)^{-2}\right)'\:$
$\to y'=3\left(-\dfrac{2}{\left(2x+5\right)^3}\cdot \:2\right)$
$\to y'=-\dfrac{12}{\left(2x+5\right)^3}$
c.Ta có:
$y=x^2\cos x$
$\to y'=(x^2\cos x)'$
$\to y'=\left(x^2\right)'\cos \left(x\right)+\left(\cos \left(x\right)\right)'\:x^2$
$\to y'=2x\cos \left(x\right)+\left(-\sin \left(x\right)\right)x^2$
$\to y'=2x\cos \left(x\right)-x^2\sin \left(x\right)$
