\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)

\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)

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

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

3/ \(y'=\dfrac{-\left(x-\sqrt{x+1}\right)'}{\left(x-\sqrt{x+1}\right)^2}=\dfrac{-1+\dfrac{1}{2\sqrt{x+1}}}{\left(x-\sqrt{x+1}\right)^2}\)

4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)

5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)

Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học

ok cảm ơn bạn nhìu

1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)

2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)

3. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)

a: \(y'=\dfrac{\left(x^2+3x-1\right)'\cdot\left(x+2\right)-\left(x^2+3x-1\right)\cdot\left(x+2\right)'}{\left(x+2\right)^2}\)

\(=\dfrac{\left(2x+3\right)\left(x+2\right)-\left(x^2+3x-1\right)}{\left(x+2\right)^2}\)

\(=\dfrac{2x^2+7x+6-x^2-3x+1}{\left(x+2\right)^2}=\dfrac{x^2+4x+7}{\left(x+2\right)^2}\)

b: \(y'=\dfrac{\left(2x^2-x\right)'\cdot\left(x^2+1\right)-\left(2x^2-x\right)\left(x^2+1\right)'}{\left(x^2+1\right)^2}\)

\(=\dfrac{4x\left(x^2+1\right)-2x\left(2x^2-x\right)}{\left(x^2+1\right)^2}\)

\(=\dfrac{4x^3+4x-4x^3+2x^2}{\left(x^2+1\right)^2}=\dfrac{2x^2+4x}{\left(x^2+1\right)^2}\)

c: \(\left(\dfrac{3-2x}{x-1}\right)'=\dfrac{\left(3-2x\right)'\left(x-1\right)-\left(3-2x\right)\left(x-1\right)'}{\left(x-1\right)^2}\)

\(=\dfrac{-2\left(x-1\right)-\left(3-2x\right)}{\left(x-1\right)^2}=\dfrac{-2x+2-3+2x}{\left(x-1\right)^2}=-\dfrac{1}{\left(x-1\right)^2}\)

\(\left(\sqrt{2x-3}\right)'=\dfrac{\left(2x-3\right)'}{2\sqrt{2x-3}}=\dfrac{1}{\sqrt{2x-3}}\)

\(y'=\left(\dfrac{3-2x}{x-1}\right)'+\left(\sqrt{2x-3}\right)'\)

\(=\dfrac{-1}{\left(x-1\right)^2}+\dfrac{1}{\sqrt{2x-3}}\)

a: \(\lim\limits_{x\rightarrow2^+}\dfrac{\sqrt{x-2}+1}{x^2-3x+2}=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2^+}\sqrt{x-2}+1=\sqrt{2-2}+1=1>0\\\lim\limits_{x\rightarrow2^+}x^2-3x+2=\lim\limits_{x\rightarrow2^+}\left(x-1\right)\left(x-2\right)=0\end{matrix}\right.\)

=>x=2 là tiệm cận đứng của đồ thị hàm số \(y=\dfrac{\sqrt{x-2}+1}{x^2-3x+2}\)

b: \(\lim\limits_{x\rightarrow-5^+}\dfrac{\sqrt{5+x}-1}{x^2+4x}=\dfrac{\sqrt{5-5}-1}{\left(-5\right)^2+4\cdot\left(-5\right)}=\dfrac{-1}{25-20}=\dfrac{-1}{5}\)

=>x=-5 không là tiệm cận đứng của đồ thị hàm số \(y=\dfrac{\sqrt{5+x}-1}{x^2+4x}\)

\(\lim\limits_{x\rightarrow\left(-4\right)^+}\dfrac{\sqrt{5+x}-1}{x^2+4x}\)

\(=\lim\limits_{x\rightarrow\left(-4\right)^+}\dfrac{5+x-1}{\left(\sqrt{5+x}+1\right)\left(x^2+4x\right)}=\lim\limits_{x\rightarrow\left(-4\right)^+}\dfrac{x+4}{\left(\sqrt{5+x}+1\right)\cdot x\left(x+4\right)}\)

\(=\lim\limits_{x\rightarrow\left(-4\right)^+}\dfrac{1}{x\left(\sqrt{5+x}+1\right)}=\dfrac{1}{\left(-4\right)\cdot\left(\sqrt{5-4}+1\right)}=\dfrac{1}{-8}=-\dfrac{1}{8}\)

=>x=-4 không là tiệm cận đứng của đồ thị hàm số \(y=\dfrac{\sqrt{5+x}-1}{x^2+4x}\)

\(\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{5+x}-1}{x^2+4x}=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow0^+}\sqrt{5+x}-1=\sqrt{5+0}-1=\sqrt{5}-1>0\\\lim\limits_{x\rightarrow0^+}x^2+4x=0\end{matrix}\right.\)

=>Đường thẳng x=0 là tiệm cận đứng của đồ thị hàm số \(y=\dfrac{\sqrt{5+x}-1}{x^2+4x}\)

c: \(\lim\limits_{x\rightarrow0^+}\dfrac{5x+1-\sqrt{x+1}}{x^2+2x}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{\dfrac{5x+1-x^2-2x-1}{5x+1+\sqrt{x+1}}}{x\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{-x^2+3x}{\left(5x+1+\sqrt{x+1}\right)\cdot x\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{-x\left(x-3\right)}{x\left(x+2\right)\left(5x+1+\sqrt{x+1}\right)}\)

\(=\lim\limits_{x\rightarrow0^+}\dfrac{-x+3}{\left(x+2\right)\left(5x+1+\sqrt{x+1}\right)}=\dfrac{-0+3}{\left(0+2\right)\left(5\cdot0+1+\sqrt{0+1}\right)}\)

\(=\dfrac{3}{2\cdot\left(6+1\right)}=\dfrac{3}{14}\)

=>x=0 không là tiệm cận đứng của đồ thị hàm số \(y=\dfrac{5x+1-\sqrt{x+1}}{x^2+2x}\)

\(\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{5x+1-\sqrt{x+1}}{x^2+2x}\) không có giá trị vì khi x=-2 thì căn x+1 vô giá trị

=>Đồ thị hàm số \(y=\dfrac{5x+1-\sqrt{x+1}}{x^2+2x}\) không có tiệm cận đứng

d: \(\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{4x^2-1}+3x^2+2}{x^2-x}\) không có giá trị vì khi x=0 thì \(\sqrt{4x^2-1}\) không có giá trị

\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{4x^2-1}+3x^2+2}{x^2-x}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1^+}\sqrt{4x^2-1}+3x^2+2=\sqrt{4-1}+3\cdot1^2+2=5+\sqrt{3}>0\\\lim\limits_{x\rightarrow1^+}x^2-x=0\end{matrix}\right.\)

=>x=1 là tiệm cận đứng của đồ thị hàm số \(y=\dfrac{\sqrt{4x^2-1}+3x^2+2}{x^2-x}\)

1) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)

\(\Rightarrow y=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)

2) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)

\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1\left(x+9\right)}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{-6}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(1\right)=\dfrac{-6}{\left(1+3\right)^2}+\dfrac{2}{\sqrt[]{1}}=-\dfrac{3}{8}+2=\dfrac{13}{8}\)

1) \(f\left(x\right)=2x-5\)

\(f'\left(x\right)=2\)

\(\Rightarrow f'\left(4\right)=2\)

2) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)

\(\Rightarrow y'=2x-\dfrac{3}{2\sqrt[]{x}}-\dfrac{1}{x^2}\)

3) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)

\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)

\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)