Exercicis derivades 01

Exercicis de derivades destinat a batxillerat.

Batxillerat

1) $f(x)={\frac {x-3}{x^{2}}}$

Solució

$f'(x)=\left({\frac {x-3}{x^{2}}}\right)'=$ ${\frac {1\cdot \left(x^{2}\right)-(x-3)\cdot 2x}{\left(x^{2}\right)^{2}}}=$ ${\frac {x^{2}-(x-3)\cdot 2x}{x^{4}}}=$ ${\frac {x-(x-3)\cdot 2}{x^{3}}}=$ ${\frac {x-2x+6}{x^{3}}}=$ ${\frac {-x+6}{x^{3}}}$ Compareu aquest altre procediment: $f'(x)=\left({\frac {x-3}{x^{2}}}\right)'=$ $\left({\frac {x}{x^{2}}}-{\frac {3}{x^{2}}}\right)'=$ $\left({\frac {1}{x}}\right)'-\left({\frac {3}{x^{2}}}\right)'=$ $-{\frac {1}{x^{2}}}+2\cdot {\frac {3}{x^{3}}}=$ $-{\frac {1}{x^{2}}}+{\frac {6}{x^{3}}}$

2) $f(x)={\frac {x^{4}-1}{x}}$

Solució

$f'(x)=\left({\frac {x^{4}-1}{x}}\right)'=$ ${\frac {\left(4x^{3}\right)\cdot x-\left(x^{4}-1\right)\cdot 1}{x^{2}}}=$ ${\frac {4x^{4}-x^{4}+1}{x^{2}}}=$ ${\frac {3x^{4}+1}{x^{2}}}$

3) $f(x)={\frac {{\sqrt {x}}+x}{x}}$

Solució

$f'(x)=\left({\frac {{\sqrt {x}}+x}{x}}\right)'=$ ${\frac {\left({\frac {1}{2{\sqrt {x}}}}+1\right)\cdot x-\left({\sqrt {x}}+x\right)\cdot 1}{x^{2}}}=$ ${\frac {{\frac {x}{2{\sqrt {x}}}}+x-{\sqrt {x}}-x}{x^{2}}}=$ ${\frac {{\frac {x\cdot {\sqrt {x}}}{2{\sqrt {x}}\cdot {\sqrt {x}}}}+x-{\sqrt {x}}-x}{x^{2}}}=$ ${\frac {{\frac {1}{2}}{\sqrt {x}}+x-{\sqrt {x}}-x}{x^{2}}}=$ ${\frac {\left({\frac {1}{2}}-1\right){\sqrt {x}}+x-x}{x^{2}}}=$ ${\frac {-{\frac {1}{2}}{\sqrt {x}}}{x^{2}}}=$ $-{\frac {1}{2}}{\frac {\sqrt {x}}{x^{2}}}$

4) $f(x)={\frac {x{\sqrt {x}}}{x^{3}}}$

Solució

$f'(x)=\left({\frac {x{\sqrt {x}}}{x^{3}}}\right)'=$ $\left({\frac {\sqrt {x}}{x^{2}}}\right)'=$ ${\frac {\left({\frac {1}{2{\sqrt {x}}}}\right)\cdot x^{2}-{\sqrt {x}}\cdot 2x}{\left(x^{2}\right)^{2}}}=$ ${\frac {{\frac {1\cdot {\sqrt {x}}}{2{\sqrt {x}}\cdot {\sqrt {x}}}}\cdot x^{2}-{\sqrt {x}}\cdot 2x}{\left(x^{2}\right)^{2}}}=$ ${\frac {{\frac {\sqrt {x}}{2x}}\cdot x^{2}-{\sqrt {x}}\cdot 2x}{x^{4}}}=$ ${\frac {{\frac {\sqrt {x}}{2}}\cdot x-{\sqrt {x}}\cdot 2x}{x^{4}}}=$ ${\frac {{\frac {\sqrt {x}}{2}}-{\sqrt {x}}\cdot 2}{x^{3}}}=$ $-{\frac {3}{2}}{\frac {\sqrt {x}}{x^{3}}}$ Compareu aquest altre procediment: $f'(x)=\left({\frac {x{\sqrt {x}}}{x^{3}}}\right)'=$ $\left(x^{1+{\frac {1}{2}}-3}\right)'=$ $\left(x^{-{\frac {3}{2}}}\right)'=$ $-{\tfrac {3}{2}}x^{-{\frac {5}{2}}}$

5) $f(x)={\frac {3+x}{x^{2}}}$

Solució

$f'(x)=\left({\frac {3+x}{x^{2}}}\right)'=$ ${\frac {(3+x)'\cdot x^{2}-(3+x)\cdot (x^{2})'}{(x^{2})^{2}}}=$ ${\frac {1\cdot x^{2}-(3+x)\cdot 2x}{x^{4}}}=$ ${\frac {x-(3+x)\cdot 2}{x^{3}}}=$ ${\frac {x-6-2x}{x^{3}}}=$ ${\frac {-x-6}{x^{3}}}$ Compareu aquest altre procediment: $f'(x)=\left({\frac {3}{x^{2}}}+{\frac {x}{x^{2}}}\right)'=$ $\left({\frac {3}{x^{2}}}\right)'+\left({\frac {1}{x}}\right)'=$ $-{\frac {6}{x^{3}}}-{\frac {1}{x^{2}}}$

6) $f(x)={\frac {1}{x^{8}}}$

Solució

$f'(x)=\left({\frac {1}{x^{8}}}\right)'=$ $-8\cdot x^{-8-1}=$ $-8\cdot x^{-9}$