Dériver un quotient de sommes de fonctions affine, carré, cube, inverse, racine carrée, puissance, de produits de fonctions affine, carré, cube, inverse, racine carrée, puissance, et de fonctions affines composées par une fonction carré, cube, inverse, racine carrée ou puissance Exercice

\forall x \in D_f' , f'(x)=\dfrac{\left( −6x+3-\dfrac{3}{x^2} \right) \sqrt{x−1}+\dfrac{\left( 3x^2−3x+3-\dfrac{3}{x} \right)}{2\sqrt{x−1}}}{x−1}

\forall x \in D_f' , f'(x)=\dfrac{\left( 1-\dfrac{1}{x^2} \right) \sqrt{x−1}+\dfrac{\left( −3\right)}{2\sqrt{x−1}}}{x−1}

\forall x \in D_f' , f'(x)=\dfrac{\left( −6x+3-\dfrac{3}{x^2} \right) \sqrt{x−1}-\dfrac{\left( 3x^2−3x+3-\dfrac{3}{x} \right)}{2\sqrt{x−1}}}{x−1}

\forall x \in D_f' , f'(x)=\dfrac{\left( −6x+3-\dfrac{3}{x^2} \right) }{x−1}

\forall x \in D_f' , f'(x) = \frac{\left(3\sqrt{x}-\frac{3}{2\sqrt{x}}+2\right)\left( x+2 \right)^2- \left(\sqrt{x}+1\right)\left(2x−3\right) \left( 2x+4 \right)}{\left( x+2 \right)^4}

\forall x \in D_f' ,  f'(x) = \frac{\left(3\sqrt{x}-\frac{3}{2\sqrt{x}}+2\right)-\left( 2x+4 \right)}{\left( x+2 \right)^4}

\forall x \in D_f' , f'(x) = \frac{\left(\frac{1}{\sqrt{x}}\right)\left( x+2 \right)^2-\left( \left(\sqrt{x}+1\right)\left(2x−3\right) \right)\left( 2x+4 \right)}{\left( x+2 \right)^4}

\forall x \in D_f' , f'(x) = \frac{\left(3\sqrt{x}-\frac{3}{2\sqrt{x}}+2\right)\left( x+2 \right)^2-\left( \left(\sqrt{x}+1\right)\left(2x−3\right) \right)\left( 2x+4 \right)}{\left( x+2 \right)^2}

\forall x \in D_f' , f'(x) =\frac{\left(3x^2−8\right)\left( 2x-1 \right)^3-6\left( x^2+3x+1 \right)\left(x−3\right)\left( 2x-1 \right)^2}{\left( 2x-1 \right)^6}

\forall x \in D_f' ,  f'(x) =\frac{\left(3x^2−8\right)\left( 2x-1 \right)^3-\left( x^2+3x+1 \right)\left(x−3\right)\left( 2x-1 \right)^2}{\left( 2x-1 \right)^6}

\forall x \in D_f' , f'(x) =\frac{\left(3x^2−8\right)\left( 2x-1 \right)^3-6\left( x^2+3x+1 \right)\left(x−3\right)\left( 2x-1 \right)^2}{\left( 2x-1 \right)^4}

\forall x \in D_f' , f'(x) =\frac{\left(3x^2−8\right)-6\left( 2x-1 \right)^2}{\left( 2x-1 \right)^6}

\forall x \in D_f' , f'(x) = \frac{\left(12x^3+2x+3-\frac{1}{x^2}\right)\left( 2\sqrt{x}+1 \right)-\left(x^3+1\right)\left(3x+\frac{1}{x}\right)\left( \frac{1}{\sqrt{x}} \right)}{\left(2\sqrt{x}+1 \right)^2}

\forall x \in D_f' ,  f'(x) = \frac{\left(12x^3+2x+3-\frac{1}{x^2}\right)-\left( \frac{1}{\sqrt{x}} \right)}{\left(2\sqrt{x}+1 \right)^2}

\forall x \in D_f' , f'(x) = \frac{\left(12x^3+2x+3-\frac{1}{x^2}\right)\left( 2\sqrt{x}+1 \right)+\left(x^3+1\right)\left(3x+\frac{1}{x}\right)\left( \frac{1}{\sqrt{x}} \right)}{\left(2\sqrt{x}+1 \right)}

\forall x \in D_f' , f'(x) = \frac{3x^2\left(3-\frac{1}{x^2}\right)\left( 2\sqrt{x}+1 \right)-\left(x^3+1\right)\left(3x+\frac{1}{x}\right)\left( \frac{1}{\sqrt{x}} \right)}{\left(2\sqrt{x}+1 \right)^2}

\forall x \in D_f' , f'(x) =\frac{\left(-7x+7\sqrt{x}+(-7x+2)\left(1-\frac{1}{2\sqrt{x}}\right)\right) \sqrt{3x+1} -\left(−7x+2\right)\left(x-\sqrt{x}\right) \frac{3}{2\sqrt{3x+1}} }{3x+1}

\forall x \in D_f' ,  f'(x) =\frac{\left(-7x+7\sqrt{x}+(-7x+2)\left(1-\frac{1}{2\sqrt{x}}\right)\right) \sqrt{3x+1} -\left(−7x+2\right)\left(x-\sqrt{x}\right) \frac{3}{2\sqrt{3x+1}} }{\sqrt{3x+1}}

\forall x \in D_f' , f'(x) =\frac{\left(-7x+7\sqrt{x}+(-7x+2)\left(1-\frac{1}{2\sqrt{x}}\right)\right) - \frac{3}{2\sqrt{3x+1}} }{3x+1}

\forall x \in D_f' , f'(x) =\frac{\left(-7x+7\sqrt{x}+(-7x+2)\left(1-\frac{1}{2\sqrt{x}}\right)\right) \sqrt{3x+1} +\left(−7x+2\right)\left(x-\sqrt{x}\right) \frac{3}{2\sqrt{3x+1}} }{3x+1}