Bài 5: Đạo hàm cấp hai

a. Ta có \(f'\left(x\right)=\ln x+x.\frac{1}{x}+\ln x\)

              \(f"\left(x\right)=\frac{1}{x}\)

              \(f'''\left(x\right)=-\frac{1}{x^2}\)

              \(f^{\left(4\right)}\left(x\right)=\frac{2}{x^3}\)

b. Tương tự ta có : 

\(f^{\left(5\right)}\left(x\right)=-\frac{2.3}{x^4}\)

\(f^{\left(6\right)}\left(x\right)=\frac{2.3.4}{x^5}\)

Từ đó suy ra \(f^{\left(5\right)}\left(x\right)=\left(-1\right)^n\frac{\left(n-2\right)!}{x^{n-1}}\) với \(n\ge2\)

Thật vậy, ta sẽ thấy công thức đúng khi n=2,3,4,......

Ta có : \(y=e^{-x}\sin x\Rightarrow\begin{cases}y'=-e^{-x}\sin x+e^{-x}\cos x=e^{-x}\left(\cos x-\sin x\right)\\y"=-e^{-x}\left(\cos x-\sin x\right)+e^{-x}\left(-\cos x-\sin x\right)=-2e^{-x}\cos x\end{cases}\)

\(\Rightarrow y"+2y'+2y=-2e^{-x}\cos x+2e^{-x}\left(\cos x-\sin x\right)+2e^{-x}\sin x=0\) => Điều phải chứng minh

Ta có : \(y=\ln\left(\frac{1}{1+x}\right)\Rightarrow y'=\frac{-\frac{1}{\left(1+x\right)^2}}{\frac{1}{1+x}}=\frac{-1}{1+x}\)

\(\Rightarrow\begin{cases}xy'+1=\frac{-x}{1+x}+1=\frac{1}{1+x}\\e^y=e^{\ln\left(\frac{1}{1+x}\right)}=\frac{1}{1+x}\end{cases}\)

\(\Rightarrow xy'+1=e^y\) (điều phải chứng minh)

Ta có : \(y=\frac{1}{1+x+\ln x}\Rightarrow y'=\frac{-\left(1+\frac{1}{x}\right)}{\left(1+x+\ln x\right)^2}=\frac{-\left(1+x\right)}{x\left(1+x+\ln x\right)^2}\)

\(\Rightarrow\begin{cases}xy'=\frac{-\left(1+x\right)}{\left(1+x+\ln x\right)^2}\\y\left(y\ln x-1\right)=\frac{1}{1+x+\ln x}\left(\frac{\ln}{1+x+\ln x}-1\right)=\frac{-\left(1+x\right)}{\left(1+x+\ln x\right)^2}\end{cases}\)

\(\Rightarrow xy'=y\left(y\ln x-1\right)\Rightarrow\) Điều phải chứng minh

Ta có : \(y=\sin\left(\ln x\right)+\cos\left(\ln x\right)\Rightarrow\begin{cases}y'=\frac{1}{x}\cos\left(\ln x\right)-\frac{1}{x}\sin\left(\ln x\right)=\frac{\cos\left(\ln x\right)-\sin\left(\ln x\right)}{x}\\y"=\frac{\left[-\frac{1}{x}\sin\left(\ln x\right)-\frac{1}{x}\cos\left(\ln x\right)\right]x-\left[\cos\left(\ln x\right)-\sin\left(\ln x\right)\right]}{x^2}=\frac{-2\cos\left(\ln x\right)}{x^2}\end{cases}\)

\(\Rightarrow y+xy'+x^2y"=\sin\left(\ln x\right)+\cos\left(\ln x\right)+\cos\left(\ln x\right)-\sin\left(\ln x\right)-2\cos\left(\ln x\right)=0\)

=> Điều cần chứng minh

Ta có \(y'=\frac{\frac{1}{x}x\left(1-\ln x\right)-\left[1-\ln x+x\left(-\frac{1}{x}\right)\right]\left(1+\ln x\right)}{x^2\left(1-\ln x\right)^2}=\frac{1-\ln x+\ln x\left(1+\ln x\right)}{x^2\left(1-\ln x\right)^2}=\frac{1+\ln^2x}{x^2\left(1-\ln x\right)^2}\)

\(\Rightarrow\begin{cases}2x^2y'=2x^2\frac{1+\ln^2x}{x^2\left(1-\ln x\right)^2}=\frac{2\left(1+\ln^2x\right)}{\left(1-\ln x\right)^2}\\x^2y^2+1=x^2\frac{1+\ln^2x}{x^2\left(1-\ln x\right)^2}+1=\frac{\left(1+\ln^2x\right)}{\left(1-\ln x\right)^2}+1=\frac{2\left(1+\ln^2x\right)}{\left(1-\ln x\right)^2}\end{cases}\)

\(\Rightarrow2x^2y'=x^2y^2+1\Rightarrow\) Điều phải chứng minh

Ta có : \(y'=x+\frac{1}{2}\left(\sqrt{x^2+1}+x\frac{x}{\sqrt{x^2+1}}\right)+\frac{\frac{1+\frac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}}{\sqrt{x+\sqrt{x^2+1}}}\)

                \(=x+\frac{2x^2+1}{2\sqrt{x^2+1}}+\frac{x+\sqrt{x^2+1}}{2\left(x+\sqrt{x^2+1}\right)\sqrt{x^2+1}}=x+\frac{2x^2+1}{2\sqrt{x^2+1}}+\frac{1}{2\sqrt{x^2+1}}\)

                \(=x+\frac{2\left(x^2+1\right)}{2\sqrt{x^2+1}}=x+\sqrt{x^2+1}\)

\(\Rightarrow\begin{cases}xy'+\ln y'=x\left(x+\sqrt{x^2+1}\right)+\ln\left(x+\sqrt{x^2+1}\right)=x^2+x\sqrt{x^2+1}+\ln\left(x+\sqrt{x^2+1}\right)\\2y=x^2+x\sqrt{x^2+1}+2\ln\sqrt{x+\sqrt{x^2+1}}=x^2+x\sqrt{x^2+1}+\ln\left(x+\sqrt{x^2+1}\right)\end{cases}\)

\(\Rightarrow2y=xy'+\ln y'\)\(\Rightarrow\) Điều phải chứng minh

\(h\left(x\right)=5\left(x+1\right)^3+4\left(x+1\right)=5\left(x^3+3x^2+3x+1\right)+4x+4=5x^3+15x^2+19x+9\\ \)

\(h'\left(x\right)=15x^2+15x+19\)

\(h"\left(x\right)=30x+15\) 

\(h"\left(x\right)=0\Leftrightarrow30x+15=0\Leftrightarrow x=\frac{-1}{2}\)

\(f'\left(x\right)=x^2+x+1\) luôn lớn hơn 0 mà :3 vậy f'(x) \(\le\)0 là k có :3