Correction – Exercice 1

Exercice 1

On considère les fonctions $f$ dérivables sur l’intervalle $I$ indiqué. Dans chacun des cas, déterminer $f'(x)$.

  1. $f(x) = -4x^2+56x-96 \quad I = \R$
    $\quad$
  2. $f(x) = (4x+7)(7x+10) \quad I= \R$
    $\quad$
  3. $f(x) = \dfrac{3 x – 4}{2x+1} \quad I=\R \backslash \left\{-\dfrac{1}{2} \right\}$
    $\quad$
  4. $f(x) = \dfrac{8+3x}{1-6x} \quad I=\R \backslash \left\{ \dfrac{1}{6} \right\}$
    $\quad$
  5. $f(x) = \dfrac{\sqrt{x}}{2 x -8} \quad I=\R^+ \backslash \{4 \}$
    $\quad$
  6. $f(x) = \dfrac{x^2+18x}{6x+4} \quad I=\R \backslash \left\{-\dfrac{2}{3} \right\}$
    $\quad$
  7. $f(x) = \dfrac{3 x -2}{2x^2-3x+5} \quad I=\R$

Correction

  1. $f'(x) = -8x + 56$
    $\quad$
  2. $f'(x)=4(7 x + 10) + 7(4 x + 7)$ $=28 x + 40 + 28 x + 49$ $=56 x +89$
    $\quad$
  3. $f'(x)= \dfrac{3(2 x + 1) – 2(3 x – 4)}{(2 x + 1)^2} $ $=\dfrac{6 x + 3 – 6x + 8}{(2 x + 1)^2}$ $=\dfrac{11}{(2 x + 1)^2}$
    $\quad$
  4. $f'(x)=\dfrac{3(1 – 6 x)-(-6)(8 + 3 x)}{(1 – 6 x)^2} $ $= \dfrac{3 – 18 x + 48 + 18 x}{(1 – 6 x)^2} $ $=\dfrac{51}{(1 – 6 x)^2}$
    $\quad$
  5. $\quad$
    $\begin{align} f'(x) &=\dfrac{\dfrac{1}{2\sqrt{x}}(2 x – 8)  -2\sqrt{x}}{(2 x – 8)^2}\\\\
    &= \dfrac{\dfrac{2 x – 8}{2} – 2x}{\sqrt{x}(2 x – 8)^2}\\\\
    &=\dfrac{x – 4- 2x}{\sqrt{x}(2 x – 8)^2} \\\\
    &=\dfrac{-x-4}{\sqrt{x}(2 x – 8)^2}
    \end{align}$
    $\quad$
  6. $\quad$
    $\begin{align} f'(x) &= \dfrac{(2x + 18)(6 x + 4) – 6(x^2 + 18x)}{(6 x + 4)^2} \\\\
    &=\dfrac{12x^2 + 8x + 108 x + 72 – 6x^2 – 108 x}{(6x + 4)^2} \\\\
    &=\dfrac{6x^2 +8x + 72}{(6x + 4)^2}
    \end{align}$
    $\quad$
  7. $\quad$
    $\begin{align} f'(x) &= \dfrac{3(2x^2 -3x + 5) – (4x – 3)(3x – 2)}{\left(2x^2- 3x + 5 \right)^2} \\\\
    &= \dfrac{6x^2 – 9x + 15 – (12x^2 – 8x – 9x + 6)}{\left(2x^2- 3x + 5 \right)^2} \\\\
    &= \dfrac{-6x^2+ 8x + 9}{\left(2x^2- 3x + 5 \right)^2}
    \end{align}$