Exercice : Exercices d'applications du 2eme tableau

Déterminer les dérivées de fonctions suivantes ainsi que les ensembles de définitions :

Question

\(a(x)=5x^3\)

Indice

\(ku \mapsto ku'\)

\(x^n \mapsto nx^{n-1}\)

Solution

\(a'(x)=5 \times 3x^2=15x^2\) \(D_a=\mathbb{R}\) \(D_{a'}=\mathbb{R}\)

Question

\(b(x)=-3x-5\)

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(x \mapsto 1\)

\(k \mapsto 0\)

Solution

\(b'(x)=-3 \times 1+0=-3\) \(D_b=\mathbb{R}\) \(D_{b'}=\mathbb{R}\)

Question

\(c(x)=10x^7-3x^4+5x+100\)

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(x^n \mapsto nx^{n-1}\)

\(k \mapsto 0\)

Solution

\(c'(x)=10 \times 7x^6-3\times 4x^3+5\times 1+0\)

\(\iff c'(x)=70x^6-12x^3+5\) \(D_c=\mathbb{R}\) \(D_{c'}=\mathbb{R}\)

Question

\(d(x)=-\frac{2}{5} x^5-πx+\frac{1}{3}x\)

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(x^n \mapsto nx^{n-1}\)

\(k \mapsto 0\)

Solution

\(d'(x)=-\frac{2}{5} \times 5x^4-π \times 1 +\frac{1}{3} \times 1\)

\(\iff d'(x)=-2x^4-π +\frac{1}{3}\)

\(D_d=\mathbb{R}\) \(D_{d'}=\mathbb{R}\)

Question

\(e(x)=3x^4-2\sqrt{x}\)

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(x^n \mapsto nx^{n-1}\)

\(\sqrt{x} \mapsto \frac{1}{2\sqrt{x}}\)

Solution

\(e'(x)=3\times 4x^3-2\frac{1}{2\sqrt{x}}\)

\(\iff e'(x)=12x^3-\frac{1}{\sqrt{x}}\)

\(D_e=[0;+\infty[\) \(D_{e'}=]0;+\infty[\)

Question

\(f(x)=3x^3+2x^2-4x+71\)

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(k \mapsto 0\)

\(x^n \mapsto nx^{n-1}\)

Solution

\(f'(x)=3 \times 3x^2+2 \times 2x-4 \times 1+0\)

\(\iff f'(x)=9x^2+4x-4\)

\(D_f=\mathbb{R}\) \(D_{f'}=\mathbb{R}\)

Question

\(g(x)=(x+1)(3-2x^2)\)

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(uv \mapsto u'v+uv'\)

\(k \mapsto 0\)

\(x^n \mapsto nx^{n-1}\)

Solution

\(g(x)=(x+1)(3-2x^2)\)

\(\begin{cases}u=x+1\\v=3-2x^2\end{cases}\)

\(\iff \begin{cases}u'=1\\v'=-2\times 2x\end{cases}\)

\(\iff \begin{cases}u'=1\\v'=-4x\end{cases}\)

\(g'(x)=1(3-2x^2)+(x+1)(-4x)\)

\(\iff g'(x)=3-2x^2+(x+1)(-4x)\)

\(\iff g'(x)=3-2x^2+(-4x)\times x+(-4x) \times 1\)

\(\iff g'(x)=3-2x^2-4x^2-4x\)

\(\iff g'(x)=-6x^2-4x+3\)

\(D_g=\mathbb{R}\) \(D_{g'}=\mathbb{R}\)

Question

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

Indice

\(ku \mapsto ku'\)

\(u+v \mapsto u'+v'\)

\(\frac{1}{x^n} \mapsto \frac{-n}{x^{n+1}}\)

Solution

\(h'(x)=\frac{-1}{x^2}-3\frac{-2}{x^3}\)

\(\iff h'(x)=\frac{-1}{x^2}+\frac{6}{x^3}\)

\(D_g=\mathbb{R}\backslash\{0\}\)\(D_{g'}=\mathbb{R}\backslash\{0\}\)

Question

\(i(x)=\frac{1}{3x^4+1}\)

Indice

\(\frac{1}{u} \mapsto \frac{-u'}{u^2}\)

\(k \mapsto 0\)

\(u+v \mapsto u'+v'\)

\(x^n \mapsto nx^{n-1}\)

\(ku \mapsto ku'\)

Solution

\(i(x)=\frac{-(3\times 4x^3+0)}{(3x^4+1)^2}\)

\(\iff i(x)=\frac{-12x^3}{(3x^4+1)^2}\)

Pour tout \(x \in \mathbb{R}\),

\(x^4\ge0\)

\(\iff 3x^4\ge0\)

\(\iff 3x^4+1\ge1\)

donc \(D_i=\mathbb{R}\)\(D_{i'}=\mathbb{R}\)

Question

\(j(x)=\frac{1-x}{3x-4}\)

Indice

\(\frac{u}{v} \mapsto \frac{u'v-uv'}{v^2}\)

\(k \mapsto 0\)

\(u+v \mapsto u'+v'\)

\(ku \mapsto ku'\)

\(x \mapsto 1\)

Solution

\(\begin{cases}u=1-x\\v=3x-4\end{cases}\)

\(\iff \begin{cases}u'=-1\\v'=3\end{cases}\)

\(j'(x)=\frac{-1(3x-4)-(1-x)(3)}{(3x-4)^2}\)

\(\iff j'(x)=\frac{-3x+4-(3-3x)}{(3x-4)^2}\)

\(\iff j'(x)=\frac{-3x+4-3+3x}{(3x-4)^2}\)

\(\iff j'(x)=\frac{1}{(3x-4)^2}\)

\(3x-4=0\)

\(\iff 3x=4\)

\(\iff x=\frac{4}{3}\)

donc \(D_j=\mathbb{R}\backslash\{\frac{4}{3}\}\)\(D_{j'}=\mathbb{R}\backslash\{\frac{4}{3}\}\)

Question

\(k(x)=\frac{\sqrt{x}}{3x^2-1}\)

Indice

\(\frac{u}{v} \mapsto \frac{u'v-uv'}{v^2}\)

\(k \mapsto 0\)

\(u+v \mapsto u'+v'\)

\(ku \mapsto ku'\)

\(x^2 \mapsto 2x\)

\(\sqrt{x} \mapsto \frac{1}{2\sqrt{x}}\)

Solution

\(\begin{cases}u=\sqrt{x}\\v=3x^2-1\end{cases}\)

\(\iff \begin{cases}u'=\frac{1}{2\sqrt{x}}\\v'=3\times 2x\end{cases}\)

\(\iff \begin{cases}u'=\frac{1}{2\sqrt{x}}\\v'=6x\end{cases}\)

\(k'(x)=\frac{\frac{1}{2\sqrt{x}} \times (3x^2-1) -\sqrt{x} \times 6x}{(3x^2-1)^2}\)

\(\iff k'(x)=\frac{1}{2\sqrt{x}} \times \frac{ (3x^2-1) -2 \sqrt{x} \times \sqrt{x} \times 6x}{(3x^2-1)^2}\)

\(\iff k'(x)=\frac{1}{2\sqrt{x}} \times \frac{ 3x^2-1 -2 x \times 6x}{(3x^2-1)^2}\)

\(\iff k'(x)=\frac{1}{2\sqrt{x}} \times \frac{ 3x^2-1 -12 x^2}{(3x^2-1)^2}\)

\(\iff k'(x)=\frac{ -9 x^2-1}{2\sqrt{x}(3x^2-1)^2}\)

\(\sqrt{x}\) existe \(si x\ge0\)

\(3x^2-1=0\)

\(\iff 3x^2=1\)

\(\iff x^2=\frac{1}{3}\)

\(\iff x_1=\frac{1}{\sqrt{3}} ou x_2=\frac{-1}{\sqrt{3}}\)

\(\iff x_1=\frac{\sqrt{3}}{\sqrt{3}^2} ou x_2=\frac{-\sqrt{3}}{\sqrt{3}^2}\)

\(\iff x_1=\frac{\sqrt{3}}{3} ou x_2=\frac{-\sqrt{3}}{3}\)

donc \(D_k=[0;+\infty[\backslash\{\frac{\sqrt{3}}{3}\}\)

\(D_{k'}=]0;+\infty[\backslash\{\frac{\sqrt{3}}{3}\}\)

Question

\(l(x)=\frac{2x^2-5x+1}{x^2+x+1}\)

Indice

\(\frac{u}{v} \mapsto \frac{u'v-uv'}{v^2}\)

\(k \mapsto 0\)

\(u+v \mapsto u'+v'\)

\(ku \mapsto ku'\)

\(x^n \mapsto nx^{n-1}\)

Solution

\(\begin{cases}u=2x^2-5x+1\\v=x^2+x+1\end{cases}\)

\(\iff \begin{cases}u'=4x-5\\v'=2x+1\end{cases}\)

\(l'(x)=\frac{(4x-5) \times (x^2+x+1) -  (2x^2-5x+1)\times (2x+1)}{(x^2+x+1)^2}\)

\((4x-5) \times (x^2+x+1) =4x \times x^2 +4x \times x +4x \times 1 +(-5) \times x^2+ (-5) \times x +(-5) \times 1\)

\(\iff (4x-5) \times (x^2+x+1) =4x^3 +4x^2 +4x -5 x^2 -5 x -5\)

\(\iff (4x-5) \times (x^2+x+1) =4x^3 -x^2 -x -5\)

\((2x^2-5x+1)\times (2x+1)=2x^2 \times 2x +2x^2 \times (-5x)+2x^2 \times 1 +(-5x) \times 2x+(-5x) \times 1+ 1 \times 2x +1 \times 1\)

\(\iff (2x^2-5x+1)\times (2x+1)=4x^3 +2x^2-10x^2-5x+ 2x +1\)

\(\iff (2x^2-5x+1)\times (2x+1)=4x^3 -8x^2-3x +1\)

\((4x-5) \times (x^2+x+1) -  (2x^2-5x+1)\times (2x+1)=(4x^3 -x^2 -x -5)-(4x^3 -8x^2-3x +1)\)

\(\iff (4x-5) \times (x^2+x+1) -  (2x^2-5x+1)\times (2x+1)=4x^3 -x^2 -x -5-4x^3 +8x^2+3x -1\)

\(\iff (4x-5) \times (x^2+x+1) -  (2x^2-5x+1)\times (2x+1)=7x^2+2x -6\)

\(l'(x)=\frac{7x^2+2x -6}{(x^2+x+1)^2}\)

\(x^2+x+1=0\)

\(\Delta=b^2-4ac=1^2-4\times 1 \times 1=1-4=-3\)

donc le dénominateur ne s’annule pas.

donc \(D_l=\mathbb{R}\)

\(D_{l'}=\mathbb{R}\)

Question

\(m(x)=\frac{x-10}{100x+1000}\)

Indice

\(\frac{u}{v} \mapsto \frac{u'v-uv'}{v^2}\)

\(k \mapsto 0\)

\(u+v \mapsto u'+v'\)

\(ku \mapsto ku'\)

\(x \mapsto 1\)

Solution

\(\begin{cases}u=x-10\\v=100x+1000\end{cases}\)

\(\iff \begin{cases}u'=1\\v'=100\end{cases}\)

\(m'(x)=\frac{1(100x+1000)-(x-10)\times 100}{(100x+1000)^2}\)

\(1(100x+1000)-(x-10)\times 100=100x+1000-(x\times 100-10\times 100)\)

\(\iff 1(100x+1000)-(x-10)\times 100=100x+1000-(100x-1000)\)

\(\iff 1(100x+1000)-(x-10)\times 100=100x+1000-100x+1000\)

\(\iff 1(100x+1000)-(x-10)\times 100=2000\)

\(m'(x)=\frac{2000}{(100x+1000)^2}\)

\(m'(x)=\frac{2000}{100^2(x+10)^2}\)

\(m'(x)=\frac{20}{100(x+10)^2}\)

\(m'(x)=\frac{1}{5(x+10)^2}\)

\(x+10=0\)

\(\iff x=-10\)

donc \(D_m=\mathbb{R}\backslash\{-10\}\)

\(D_{m'}=\mathbb{R}\backslash\{-10\}\)

Question

\(o(x)=\frac{x-3}{x}\)

Indice

\(\frac{u}{v} \mapsto \frac{u'v-uv'}{v^2}\)

\(k \mapsto 0\)

\(u+v \mapsto u'+v'\)

\(ku \mapsto ku'\)

\(x \mapsto 1\)

Solution

\(\begin{cases}u=x-3\\v=x\end{cases}\)

\(\iff \begin{cases}u'=1\\v'=1\end{cases}\)

\(o'(x)=\frac{1\times x-(x-3)\times 1}{x^2}\)

\(o'(x)=\frac{ x-x+3}{x^2}\)

\(o'(x)=\frac{3}{x^2}\)

donc \(D_m=\mathbb{R}^*\)

\(D_{m'}=\mathbb{R}^*\)