2024/G2/12

$\mathbf {3.1}$
1. The derivative of a constant is 0 = ${\frac {d}{dx}}[c]=0$
2. ${\frac {d}{dx}}[x^{n}]=nx^{n-1}$

3. ${\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]$

4. ${\frac {d}{dx}}[f(x)+g(x)]={\frac {d}{dx}}[f(x)]+{\frac {d}{dx}}[g(x)]$

5. ${\frac {d}{dx}}[a^{x}]=ln(a)\cdot a^{x}$

6. ${\frac {d}{dx}}[e^{x}]=e^{x}$


$\mathbf {\color {Blue}{Examples}}$

1) ${\frac {d}{dx}}[5+\pi ]=0+0=0$

2) ${\frac {d}{dx}}[3x]=3\cdot {\frac {d}{dx}}[x]=3(1)(x)^{1-1}=3\cdot x^{0}=3$

3) ${\frac {d}{dx}}[9x^{2}]=18x$

4) ${\frac {d}{dx}}[x^{3}+x^{2}+10]=3x^{2}+2x+0$
5) Failed to parse (syntax error): {\displaystyle {\frac{d}{dx}} [\sqrt(x)] = {\frac{d}{dx}} [x^{{\frac{1}{2}}] = {\frac{1}{2x}^{{\frac{1}{2}-1}} }

\sqrt{x}

${\frac {d}{dx}}[c]=0$
${\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]$
${\frac {d}{dx}}[f(x)\pm g(x)]={\frac {d}{dx}}[f(x)]\pm {\frac {d}{dx}}[g(x)]$
${\frac {d}{dx}}[a^{x}]=\ln(a)a^{x}$
${\frac {d}{dx}}[e^{x}]=e^{x}$
$\color {Blue}Power\,Rule$
${\frac {d}{dx}}[x^{n}]=n\cdot x^{n}-1$
$\color {Red}Product\,Rule$
${\frac {d}{dx}}[f\cdot {g}]={\frac {d}{dx}}[f]\cdot {g}+{\frac {d}{dx}}[g]\cdot {f}$
$\color {Green}Quotient\,Rule$
${\frac {d}{dx}}[{\frac {f}{g}}]={\frac {{\frac {d}{dx}}[f]\cdot {g}-{\frac {d}{dx}}[g]\cdot {f}}{g^{2}}}$
$\mathbf {\color {Purple}{Examples}}$
$\mathbf {Ex.1}$
$if\,f(x)=x\cdot {e^{x}}$
$f^{\prime }(x)=1\cdot {e^{x}}+x\cdot {e^{x}}$
$\mathbf {Ex.2}$
$if\,f(t)={\sqrt {t}}(a+bt)$
$f^{\prime }(t)={\frac {1}{2{\sqrt {t}}}}(a+bt)+t{\sqrt {t}}(b)$
$\mathbf {Ex.3}$
$if\,f(x)={\sqrt {x}}\cdot {g(x)}$
$g(4)=2$
$g^{\prime }(4)=3$
$f^{\prime }(x)={\frac {1}{2{\sqrt {x}}}}\cdot {g(x)}+{\sqrt {x}}\cdot {g^{\prime }(x)}$
$\mathbf {Ex.4}$
$y={\frac {\color {Blue}{x^{2}+x-2}}{\color {Red}{x^{3}+6}}}$
${\frac {d}{dx}}=y^{\prime }={\frac {(2x+1)(x^{3}-6)-\color {Blue}{(x^{2}+x-2)}(3x^{2})}{\color {Red}{(x^{3}+6)^{2}}}}$
$={\frac {(2x^{4}+x^{4}+x^{3}+12x+6-[3x^{4}+3x^{2}-6x^{2}]}{(x^{3}+6)^{2}}}$
$={\frac {-x^{4}-2x^{3}+6x^{2}+12x+6}{(x^{3}+6)^{2}}}$
$\mathbf {Ex.5}$
$y={\frac {e^{x}}{1+x^{2}}}\,(1,{\frac {e}{2}})\,$
${\frac {d}{dx}}={\frac {e^{x}\cdot (1+x^{2})-e^{x}(2x)}{(1+x^{2})^{2}}}$
${\frac {d}{dx}}|_{x=1}{\frac {e(1+1)-e^{\prime }(2)}{(1+1)^{2}}}={\frac {2e-2e}{2^{2}}}={\frac {0}{4}}=0$