Exercicis de derivades destinat a batxillerat.

### Batxillerat

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

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

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

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

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

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

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

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

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

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

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

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