Tìm đạo hàm của hàm số \(y=\frac{1}{4\left(1+x^4\right)}+\frac{1}{4}\ln\left(\frac{x^4}{1+x^4}\right)\) .
\(\dfrac{1}{x\left(1+x^4\right)^2}\) \(\dfrac{1}{x^2\left(1+x^4\right)^2}\) \(\dfrac{1}{x^3\left(1+x^4\right)^2}\) \(\dfrac{1}{x^4\left(1+x^4\right)}\) Hướng dẫn giải:Ta có \(\left(\dfrac{1}{\left(1+x^4\right)}\right)'=\dfrac{-4x^3}{\left(1+x^4\right)^2}=-\dfrac{4x^3}{\left(1+x^4\right)^2}\Rightarrow\left(\dfrac{1}{4\left(1+x^4\right)}\right)'=-\dfrac{x^3}{\left(1+x^4\right)^2}\)
và \(\dfrac{x^4}{1+x^4}=1-\dfrac{1}{1+x^4}\Rightarrow\left(\dfrac{x^4}{1+x^4}\right)'=\left(-\dfrac{1}{1+x^4}\right)'=\dfrac{4x^3}{\left(1+x^4\right)^2}\) suy ra
\(\left(\ln\left(\dfrac{x^4}{1+x^4}\right)\right)'=\left(\dfrac{x^4}{1+x^4}\right)':\left(\dfrac{x^4}{1+x^4}\right)=\dfrac{4x^3}{\left(1+x^4\right)^2}.\dfrac{1+x^4}{x^4}=\dfrac{4}{\left(1+x^4\right)x}\)\(\Rightarrow\left(\dfrac{1}{4}\ln\left(\dfrac{x^4}{1+x^4}\right)\right)'=\dfrac{1}{x\left(1+x^4\right)}\)
Do đó \(y'=-\dfrac{x^3}{\left(1+x^4\right)^2}+\dfrac{1}{x\left(1+x^4\right)}=\dfrac{-x^4+\left(1+x^4\right)}{x\left(1+x^4\right)^2}=\dfrac{1}{x\left(1+x^4\right)^2}\)