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

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

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

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

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

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

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

a: \(y'=\left(x^2+2x\right)'\left(x^3-3x\right)+\left(x^2+2x\right)\left(x^3-3x\right)'\)

\(=\left(2x+2\right)\left(x^3-3x\right)+\left(x^2+2x\right)\left(3x^2-3\right)\)

\(=2x^4-6x^2+2x^3-6x+3x^4-3x^2+6x^3-6x\)

\(=5x^4+8x^3-9x^2-12x\)

b: y=1/-2x+5 

=>\(y'=\dfrac{2}{\left(2x+5\right)^2}\)

c: \(y'=\dfrac{\left(4x+5\right)'}{2\sqrt{4x+5}}=\dfrac{4}{2\sqrt{4x+5}}=\dfrac{2}{\sqrt{4x+5}}\)

d: \(y'=\left(sinx\right)'\cdot cosx+\left(sinx\right)\cdot\left(cosx\right)'\)

\(=cos^2x-sin^2x=cos2x\)

e: \(y=x\cdot e^x\)

=>\(y'=e^x+x\cdot e^x\)

f: \(y=ln^2x\)

=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)

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)

a) Với x > 0 bất kì và \(h = x - {x_0}\) ta có

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {{x_0} + h} \right) - \ln {x_0}}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}.{x_0}}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x_0}}}.\mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{{{x_0}}}} \right)}}{{\frac{h}{{{x_0}}}}} = \frac{1}{{{x_0}}}\end{array}\)

Vậy hàm số \(y = \ln x\) có đạo hàm là hàm số \(y' = \frac{1}{x}\)

b) Ta có \({\log _a}x = \frac{{\ln x}}{{\ln a}}\) nên \(\left( {{{\log }_a}x} \right)' = \left( {\frac{{\ln x}}{{\ln a}}} \right)' = \frac{1}{{x\ln a}}\)

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

Ta có :

\(y'=\sin x+x\cos x\)

\(y"=\cos x+\cos x-x\sin x=2\cos x-x\sin x\)

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