Exercice : Polynômes 2

Calculer les dérivées respectives des fonctions suivantes et donner leurs ensembles de définitions :

Question

\(a(x)= 3x^2 − 7x + 2.\)

Indice

\(x \mapsto 1\)

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

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

\(ku\mapsto ku'\)

Solution

\(a'(x)= 3\times 2x − 7 \times 1+ 0\)

\(\iff a'(x)= 6x − 7\)

Question

\(b( x ) = x^3 -5 x^2 +x+4\)

Indice

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

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

\(b'( x ) =3 x^2 -5 \times 2x +1=3 x^2 -10x +1\)

\(D_b=\mathbb{R}\) et \(D_{b'}=\mathbb{R}\)

Question

\(c( x ) = x^3 + x^2 +x+1\)

Indice

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

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

\(c'(x)=3x^{2}+2x+1\)

\(D_c=\mathbb{R}\) et \(D_{c'}=\mathbb{R}\)

Question

\(d\left(x\right)=\frac{x^{3}}{6}-4x^{2}+2\)

Indice

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

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

\(d'\left(x\right)=\frac{1}{6}\times 3x^{2}-4\times 2x+0\)

\(d'\left(x\right)=\frac{1}{2}x^{2}-8x\)

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

Question

\(e\left(t\right)=-5t^{2}+2t\)

Indice

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

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

\(e'\left(t\right)=-5\times 2t+2=-10t+2\)

\(D_e=\mathbb{R}\) et \(D_{e'}=\mathbb{R}\)

Question

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

Indice

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

\(x \mapsto 1\)

\(\frac{1}{x} \mapsto \frac{-1}{x^2}\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

\(f'(x)=3 \times 2x-5- \frac{1}{x^2}\)

\(\iff f'(x)=6x-5- \frac{1}{x^2}\)

\(\iff f'(x)= \frac{6x^3-5x^2-1}{x^2}\)

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

Question

\(g( x ) =2x+3+ \frac{1}{2x+3}\)

Indice

\(x \mapsto 1\)

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

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

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

\(u=2x+3\iff u'=2\)

\(2x+3=0\)

\(\iff 2x=-3\)

\(\iff x=-1,5\)

\(D_g=\mathbb{R}\backslash\{-1,5\}\) et \(D_{g'}=\mathbb{R}\backslash\{-1,5\}\)

Question

\(h( t ) =\frac{-5}{t}\)

Indice

\(\frac{1}{x} \mapsto \frac{-1}{x^2}\)

\(ku\mapsto ku'\)

Solution

\(h'( t) =-5 \times \frac{-1}{t^2}\)

\(h'( t) =\frac{5}{t^2}\)

\(D_h=\mathbb{R}^*\) et \(D_{h'}=\mathbb{R}^*\)

Question

\(i( t ) =\frac{-5}{t}\)

Indice

\(\frac{1}{x} \mapsto \frac{-1}{x^2}\)

\(ku\mapsto ku'\)

Solution

\(i'( t) =-5 \times \frac{-1}{t^2}\)

\(i'( t) =\frac{5}{t^2}\)

\(D_i=\mathbb{R}^*\) et \(D_{i'}=\mathbb{R}^*\)

Question

\(j( x ) =\frac{3x+5}{2x+1}\)

Indice

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

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

Solution

\(\begin{cases}u=3x+5\\v=2x+1\end{cases}\)

\(\begin{cases}u'=3\\v'=2\end{cases}\)

\(j'( x ) =\frac{3(2x+1)-2(3x+5)}{(2x+1)^2}\)

\(\iff j'( x ) =\frac{6x+3-6x-10}{(2x+1)^2}\)

\(\iff j'( x ) =\frac{-7}{(2x+1)^2}\)

\(2x+1=0\)

\(\iff 2x=-1\)

\(\iff x=-\frac{1}{2}\)

\(D_j=\mathbb{R}\backslash\{-\frac{1}{2}\}\) et \(D_{j'}=\mathbb{R}\backslash\{-\frac{1}{2}\}\)

Question

\(k( t ) =\frac{3x-4}{x+2}\)

Indice

\(x \mapsto 1\)

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

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

\(ku\mapsto ku'\)

Solution

\(\begin{cases}u=3x-4\\v=x+2\end{cases}\)

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

\(k'( t ) =\frac{3(x+2)-(3x-4)\times 1}{(x+2)^2}\)

\(\iff k'( t) =\frac{3x+6-3x+4}{(x+2)^2}\)

\(x+2=0\)

\(\iff x=-2\)

\(D_k=\mathbb{R}\backslash\{-2\}\) et \(D_{k'}=\mathbb{R}\backslash\{-2\}\)

Question

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

Indice

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

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

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

Solution

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

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

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

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

\(\iff l'(x)=\frac{(4x^3+4x-4x^2-4)-4x^{3}+8x^2-4x}{(x^{2}+1)^2}\)

\(\iff l'(x)=\frac{4x^3+4x-4x^2-4-4x^{3}+8x^2-4x}{(x^{2}+1)^2}\)

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

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

\(x^{2}+1\) est une forme factorisée qui nous permet de savoir que le sommet de cette parabole est le point de coordonnées (0;1)

donc le dénominateur ne s’annule pas.

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

Question

\(m(x)=x^{2021}-\frac{2021}{x}+2021\sqrt{x}\)

Indice

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

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

\(\frac{1}{x} \mapsto \frac{-1}{x^2}\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

\(m'(x)=2021x^{2020}-2021\times \frac{-1}{x^2}+2021\times \frac{1}{2\sqrt{x}}\)

\(\iff m'(x)=2021x^{2020}+\frac{2021}{x^2}+\frac{2021}{2\sqrt{x}}\)

\(D_m=]0 ;+\infty[\) et \(D_{m'}=]0;+\infty[\)

Question

\(n( x ) =2x+3+ \frac{1}{2x+3}\)

Indice

\(x \mapsto 1\)

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

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

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

\(u=2x+3\iff u'=2\)

\(2x+3=0\)

\(\iff 2x=-3\)

\(\iff x=-1,5\)

\(D_n=\mathbb{R}\backslash\{-1,5\}\) et \(D_{n'}=\mathbb{R}\backslash\{-1,5\}\)

Question

\(o( x ) =\frac{3x+5}{2x+1}\)

Indice

\(x \mapsto 1\)

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

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

Solution

\(\begin{cases}u=3x+5\\v=2x+1\end{cases}\)

\(\begin{cases}u'=3\\v'=2\end{cases}\)

\(o'( x ) =\frac{3(2x+1)-2(3x+5)}{(2x+1)^2}\)

\(\iff o'( x ) =\frac{6x+3-6x-10}{(2x+1)^2}\)

\(\iff o'( x ) =\frac{-7}{(2x+1)^2}\)

\(2x+1=0\)

\(\iff 2x=-1\)

\(\iff x=-\frac{1}{2}\)

\(D_o=\mathbb{R}\backslash\{-\frac{1}{2}\}\) et \(D_{o'}=\mathbb{R}\backslash\{-\frac{1}{2}\}\)

Question

\(p( x ) =5x^3 + \frac{1}{x}\)

Indice

\(x \mapsto 1\)

\(\frac{1}{x} \mapsto \frac{-1}{x^2}\)

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

\(k\mapsto 0\)

\(ku\mapsto ku'\)

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

Solution

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

\(\iff p'( x ) =15x^2 - \frac{1}{x^2}\)

\(\iff p'( x ) =\frac{15x^4 - 1}{x^2}\)

\(D_p=\mathbb{R}^*\) et \(D_{p'}=\mathbb{R}^*\)

Question

\(q( t ) =\frac{3x-4}{x+2}\)

Indice

\(x \mapsto 1\)

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

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

\(ku\mapsto ku'\)

Solution

\(\begin{cases}u=3x-4\\v=x+2\end{cases}\)

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

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

\(\iff q'( x) =\frac{3x+6-3x+4}{(x+2)^2}\)

\(x+2=0\)

\(\iff x=-2\)

\(D_q=\mathbb{R}\backslash\{-2\}\) et \(D_{q'}=\mathbb{R}\backslash\{-2\}\)

Question

\(r(s) = \frac{s + 2}{s^2 + 1}\)

Indice

\(x \mapsto 1\)

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

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

\(ku\mapsto ku'\)

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

Solution

\(\begin{cases}u=s+2\\v=s^2 + 1\end{cases}\)

\(\begin{cases}u'=1\\v'=2s\end{cases}\)

\(r'(s) = \frac{1(s^2 + 1)-(s+2) \times 2s}{(s^2 + 1)^2}\)

\(\iff r'(s) = \frac{s^2 + 1-(2s^2+4s)}{(s^2 + 1)^2}\)

\(\iff r'(s) = \frac{s^2 + 1-2s^2-4s}{(s^2 + 1)^2}\)

\(\iff r'(s) = \frac{-s^2-4s+1}{(s^2 + 1)^2}\)

\(s^2+1\) est une forme canonique donc le sommet de la parabole d'équation \(s^2+1\) est (0;1)

donc pour tout \(s\in\mathbb{R}\) \(s^2+1\ge1\)

L'équation \(s^2 + 1=0\) n'admet donc pas de solution.

\(D_r=\mathbb{R}\) et \(D_{r'}=\mathbb{R}\)