Giải thích các bước giải:
Ta có:
$y=\sin^2\sqrt{x^2+4x+5}$
$\to y=\dfrac12\left(1-\cos\left(2\sqrt{x^2+4x+5}\right)\right)$
$\to y=\dfrac12-\dfrac12\cos\left(2\sqrt{x^2+4x+5}\right)$
$\to y'=\left(\dfrac12-\dfrac12\cos\left(2\sqrt{x^2+4x+5}\right)\right)'$
$\to y'=-\dfrac12\left(\cos\left(2\sqrt{x^2+4x+5}\right)\right)'$
$\to y'=-\dfrac12\left(-\sin \left(2\sqrt{x^2+4x+5}\right)\left(2\sqrt{x^2+4x+5}\right)'\right)$
$\to y'=-\dfrac12\left(-\sin \left(2\sqrt{x^2+4x+5}\right)\dfrac{2x+4}{\sqrt{x^2+4x+5}}\right)$
$\to y'=\dfrac12\sin \left(2\sqrt{x^2+4x+5}\right)\dfrac{2x+4}{\sqrt{x^2+4x+5}}$
