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a) Với bất kì \({x_0} \in \mathbb{R}\), ta có:
\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x - {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} 1 = 1\)
Vậy \(f'\left( x \right) = {\left( x \right)^\prime } = 1\) trên \(\mathbb{R}\).
b) Ta có:
\(\begin{array}{l}{\left( {{x^2}} \right)^\prime } = 2{\rm{x}}\\{\left( {{x^3}} \right)^\prime } = 3{{\rm{x}}^2}\\...\\{\left( {{x^n}} \right)^\prime } = n{{\rm{x}}^{n - 1}}\end{array}\)
a) \(g'\left( x \right) = y' = {\left( {2x + \frac{\pi }{4}} \right)^,}.\cos \left( {2x + \frac{\pi }{4}} \right) = 2\cos \left( {2x + \frac{\pi }{4}} \right)\)
b) \(g'\left( x \right) = - 2{\left( {2x + \frac{\pi }{4}} \right)^,}.\sin \left( {2x + \frac{\pi }{4}} \right) = - 4\sin \left( {2x + \frac{\pi }{4}} \right)\)
a: Nếu a là số nguyên dương thì TXĐ là D=R
Nếu a là số không phải nguyên dương thì TXĐ là D=R\{0}
Nếu a không là số nguyên thì TXĐ: D=R
b: \(y'=\left(x^a\right)'=\left(e^{a\cdot lnx}\right)'\)
\(=\dfrac{a}{x}\cdot e^{a\cdot lnx}=\dfrac{a}{x}\cdot x^a=a\cdot x^{a-1}\)
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. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)
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) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)
\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1.\left(x+9\right)}{\left(x-3\right)^2}+\dfrac{4}{2\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=\dfrac{12}{\left(x-3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)
\(\Rightarrow f'\left(x\right)=2\left[\dfrac{6}{\left(x-3\right)^2}+\dfrac{1}{\sqrt[]{x}}\right]\)
\(\Rightarrow f'\left(1\right)=2\left[\dfrac{6}{\left(1-3\right)^2}+\dfrac{1}{\sqrt[]{1}}\right]=2\left(\dfrac{3}{2}+1\right)=2.\dfrac{5}{2}=5\)
\(a,y'=\left(\dfrac{1}{2x+3}\right)'=-\dfrac{2}{\left(2x+3\right)^2}\\ \Rightarrow y''=\dfrac{2\cdot\left[\left(2x+3\right)^2\right]'}{\left(2x+3\right)^4}=\dfrac{8}{\left(2x+3\right)^3}\\ b,y'=\left(log_3x\right)'=\dfrac{1}{xln3}\\ \Rightarrow y''=-\dfrac{1}{x^2ln3}\\ c,y'=\left(2^x\right)'=2^x\cdot ln2\\ \Rightarrow y''=2^x\cdot\left(ln2\right)^2\)
\(a,y'=\left(x^3-4x^2+5\right)'=3x^2-8x\\ b,y''=\left(3x^2-8x\right)'=6x-8\)