Chương 5: ĐẠO HÀM

ta có \(\ln\frac{1}{1+x}=-\ln\left(1+x\right)\Rightarrow y'=-\frac{1}{1+x}\)

vậy xy'+1=\(\frac{-x}{1+x}+1=\frac{1}{1+x}=e^y\)

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

           \(xy'=\left(1-x^2\right)xe^{-\frac{x^2}{2}}=\left(1-x^2\right)y\)

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

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

Lại có \(y\left(y\ln x-1\right)=\frac{-1-x}{\left(1+x+\ln x\right)^2}\)   (2)

Từ (1) và (2) suy ra \(xy'=y\left(y\ln x-1\right)\)

Ta có \(y'=-a.e^{-x}-2b.e^{-2x};y"=ae^{-x}+4be^{-2x}\)

Vậy \(y"+3y+2y=ae^{-x}+4be^{-2x}-3ae^{-x}-6be^{-2x}+2ae^{-x}+2be^{-2x}=0\)

ta có y'=\(-e^{-x}.\sin+e^{-x}.cosx\)

y"=\(e^{-x}.sinx-e^{-x}.cosx-e^{-x}.cosx-e^{-x}.sinx=-2e^{-x.cosx}\)

vậy y"+2y'+2y=\(-2e^{-x}.cosx-2e^{-x}.sinx+2e^{-x}.cosx+2e^{-x}.sinx=0\)

ta có y'=\(e^{sinx}.\cos sx;y"=e^{sinx}.cos^2x-sinx.e^{sinx}\)

vậy y'cosx-ysinx-y"=\(e^{sinx}.cos^2x-e^{sinx}.sinx-é^{sinx}.sinx-e^{sinx}.cos^2x+e^{sinx}.sinx=0\)

Ta có \(y'=2e^{2x}\sin5x+5e^{2x}\cos5x\)

          \(y"=4e^{2x}\sin5x+10e^{2x}\cos5x+10e^{2x}\cos5x-25e^{2x}\sin5x\)

               \(=-21e^{2x}\sin5x+20e^{2x}\cos5x\)

Vậy \(y"-4y'+29=-21e^{2x}\sin5x+20e^{2x}\cos5x-8e^{2x}\cos5x+29e^{2x}\sin5x=0\)

\(f\left(x\right)=\left(x+3\right)\sqrt{9-x^2}\Rightarrow y^2=\left(x^2+6x+9\right)\left(9-x^2\right)=54x+81-x^4-6x^3\)

\(y'=\dfrac{-4x^3-18x^2+54}{2\left(x+3\right)\sqrt{9-x^2}}=\dfrac{-2x^3-48+27}{\left(x+3\right)\sqrt{9-x^2}}\)

=\(\sqrt{2-x}\) + x.\(\frac{-1}{2.\sqrt{2-x}}\)

\(y=x\sqrt{2-x}=\sqrt{2x^2-x^3}\)

\(y^2=2x^2-x^3\)

\(2y=4x-3x^2\)

\(y'=\dfrac{4x-3x^2}{2y}=\dfrac{4x-3x^2}{2x\sqrt{2-x}}=\dfrac{4-3x}{2\sqrt{2-x}}\)