a) Với \({x_0}\) bất kì, ta có:

\(\begin{array}{l}f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^3} + {x^2} - x_0^3 - x_0^2}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x - {x_0}} \right)\left( {{x^2} + x{x_0} + x_0^2} \right) + \left( {x - {x_0}} \right)\left( {x + {x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x - {x_0}} \right)\left( {{x^2} + x{x_0} + x_0^2 + x + {x_0}} \right)}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \left( {{x^2} + x{x_0} + x_0^2 + x + {x_0}} \right) = 3x_0^2 + 2{x_0}\end{array}\)

Vậy hàm số \(y = {x^3} + {x^2}\) có đạo hàm là hàm số \(y' = 3{x^2} + 2x\)

b) \({\left( {{x^3}} \right)^,} + {\left( {{x^2}} \right)^,} = 3{x^2} + 2x\)

Do đó \(\left( {{x^3} + {x^2}} \right)'\) = \(\left( {{x^3}} \right)' + \left( {{x^2}} \right)'.\)

Xét \(\Delta x\) là số gia của biến số tại điểm x

Ta có:

\(\begin{array}{l}\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = {\left( {x + \Delta x} \right)^3} - {x^3} = \left( {x + \Delta x - x} \right)\left[ {x{{\left( {x + \Delta x} \right)}^2} + x.\left( {x + \Delta x} \right) + {x^2}} \right]\\ = \Delta x\left( {{x^2} + 2x.\Delta x + {{\left( {\Delta x} \right)}^2} + {x^2} + x.\Delta x + {x^2}} \right) = \Delta x.\left( {3{x^2} + {{\left( {\Delta x} \right)}^2} + 3x.\Delta x} \right)\\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = 3{x^2} + {\left( {\Delta x} \right)^2} + 3x.\Delta x\end{array}\)

Ta thấy:

\(\begin{array}{l}\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \left( {3{x^2} + {{\left( {\Delta x} \right)}^2} + 3x.\Delta x} \right) = 3{x^2}\\ \Rightarrow f'\left( x \right) = 3{x^2}\end{array}\)

\(f'\left(3\right)=\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-f\left(3\right)}{x-3}\\ =\lim\limits_{x\rightarrow3}\dfrac{2x-6}{x-3}\\ =2\)

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

\(b,y'=\left(\sqrt{x}+1\right)'\left(x^2+2\right)+\left(\sqrt{x}+1\right)\left(x^2+2\right)'\\ =\dfrac{x^2+2}{2\sqrt{x}}+\left(\sqrt{x}+1\right)\cdot2x\)

a: \(f'\left(x_0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x0\right)}{x-x0}=\lim\limits_{x\rightarrow x0}\dfrac{c-c}{x-x0}=0\)

b: \(f'\left(x0\right)=\lim\limits_{x\rightarrow x0}\dfrac{f\left(x\right)-f\left(x0\right)}{x-x0}=\lim\limits_{x\rightarrow x0}\dfrac{x-x0}{x-x0}=1\)

a, \(y=\left(2x^3+3\right)^2\)

\(y'=2\left(2x^3+3\right)6x^2\)

\(=12x^2\left(2x^3+3\right)\)

b,\(y=cos3x\)

\(y'=-3sin3x\)

c, \(y=log_2\left(x^2+2\right)\)

\(y'=\dfrac{2x}{\left(x^2+2\right)ln2}\)

\(h\left(x\right)=f\left(x^2+1\right)-m\Rightarrow h'\left(x\right)=2x.f'\left(x^2+1\right)\)

\(h'\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\f'\left(x^2+1\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^2+1=2\\x^2+1=5\end{matrix}\right.\) \(\Rightarrow x=\left\{-2;-1;0;1;2\right\}\)

Hàm có nhiều cực trị nhất khi \(h\left(x\right)=m\) có nhiều nghiệm nhất

\(f\left(x\right)=\int f\left(x\right)dx=\dfrac{1}{4}x^4-\dfrac{5}{3}x^3-2x^2+20x+C\)

\(f\left(1\right)=0\Rightarrow C=-\dfrac{199}{12}\Rightarrow f\left(x\right)=-\dfrac{1}{4}x^4-\dfrac{5}{3}x^3-2x^2+20x-\dfrac{199}{12}\)

\(x=\pm2\Rightarrow x^2+1=5\Rightarrow f\left(5\right)\approx-18,6\)

\(x=\pm1\Rightarrow x^2+1=2\Rightarrow f\left(2\right)\approx6,1\)

\(x=0\Rightarrow x^2+1=1\Rightarrow f\left(1\right)=0\)

Từ đó ta phác thảo BBT của \(f\left(x^2+1\right)\) có dạng:

Từ đó ta dễ dàng thấy được pt \(f\left(x^2+1\right)=m\) có nhiều nghiệm nhất khi \(0< m< 6,1\)

\(\Rightarrow\) Có 6 giá trị nguyên của m

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

\(=e^x\left(2x+3\right)+\left(x^2+3x-1\right)\cdot e^x\)

\(=e^x\left(x^2+5x+2\right)\)

b: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)

\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)