Giải thích các bước giải:
a.Ta có:
$y=\dfrac{x^2+1}{\tan x}$
$\to y'=( \dfrac{x^2+1}{\tan x})'$
$\to y'=\dfrac{\left(x^2+1\right)'\tan \left(x\right)-\left(\tan \left(x\right)\right)'\left(x^2+1\right)}{\left(\tan \left(x\right)\right)^2}$
$\to y'=\dfrac{2x\tan \left(x\right)-\dfrac{1}{\cos^2x}\cdot\left(x^2+1\right)}{\left(\tan \left(x\right)\right)^2}$
$\to y'=\dfrac{2x\cos ^2\left(x\right)\tan \left(x\right)-x^2-1}{\sin ^2\left(x\right)}$
b.Ta có:
$y=\cos(x^3+\dfrac{\pi}{4})$
$\to y'=(\cos(x^3+\dfrac{\pi}{4}))'$
$\to y'=-\sin \left(x^3+\dfrac{\pi }{4}\right)\left(x^3+\dfrac{\pi }{4}\right)'\:$
$\to y'=-\sin \left(x^3+\dfrac{\pi }{4}\right)\cdot \:3x^2$
c.Ta có:
$y=\tan(\sqrt{3x+5})$
$\to y'=(\tan(\sqrt{3x+5}))'$
$\to y'=\dfrac{1}{\cos^2\left(\sqrt{3x+5}\right)}\cdot\left(\sqrt{3x+5}\right)'\:$
$\to y'=\dfrac{1}{\cos^2\left(\sqrt{3x+5}\right)}\cdot \dfrac{3}{2\sqrt{3x+5}}$
