Dérivées et intégrales

Formules

Dérivées

Attention à ajouter la constante d'intégration \( C \)

\(a, b, c, d, e \) sont des paramètres (même \(e\))
Primitive Fonction Dérivée
\[ \frac{a\blue{x}^{b + 1}}{b + 1} + c\blue{x} \] \[ a \blue{x}^b + c \] \[ b a \blue{x}^{b - 1} \]
\[ \frac{a b^{c \blue{x} + d}}{c ~ ln(b)} + e \blue{x}\] \[ ab^{c\blue{x} + d} + e \] \[ cab^{c\blue{x} +d} ln(a) \]
\[ \frac{\blue{x} log_a(\blue{x})}{ln(a)} \] \[ log_a( \blue{x} ) \] \[ \frac{1}{ \blue{x} ln(a)} \]
\[ \] \[ \blue{x}^{\blue{x}} \] \[ \blue{x}^{\blue{x}}ln(\blue{x} + 1) \]
Fonction Dérivée
\[ a \cdot f(\blue{x}) \] \[ a \cdot f'(\blue{x}) \]
\[ f(\blue{x} )\green±g(\blue{x} )\] \[ f'(\blue{x} )\green±g'(\blue{x} )\]
\[ f(\blue{x}) \cdot g(\blue{x}) \] \[ f'(\blue{x}) \cdot g(\blue{x}) + f(\blue{x}) \cdot g'(\blue{x} )\]
\[ \frac{f(\blue{x})}{g(\blue{x})} \] \[ \frac{f'(\blue{x}) g(\blue{x}) - f(\blue{x}) g'(\blue{x})}{(g(\blue{x}))^2} \]
\[ f(g(h(\blue{x}))) \] \[ f'(g(h(\blue{x}))) \cdot g'(h(\blue{x})) \cdot h'(\blue{x}) \]
\[ f^{-1}(\blue{x}) \] \[ \frac{1}{f'(f^{-1}(\blue{x}))} \]
Fonction Dérivée Fonction Dérivée
\[ sin(\blue{x}) \] \[ cos(\blue{x}) \] \[ cos(\blue{x}) \] \[ -sin(\blue{x} )\]
\[ tan(\blue{x}) \] \[ 1 + tan^2(\blue{x}) \] \[ \frac{1}{cos^2(\blue{x})} \] \[ cot(\blue{x}) \] \[ -(1 + cot^2(\blue{x})) \]
\[ sec(\blue{x}) \] \[ sec(\blue{x})tan(\blue{x}) \] \[ csc(\blue{x}) \] \[ -csc(\blue{x})cot(\blue{x}) \]
\[ sin^{-1}(\blue{x}) \] \[ \frac{1}{\sqrt{1 - x^2}} \] \[ cos^{-1}(\blue{x}) \] \[ -\frac{1}{\sqrt{1 - x^2}} \]
\[ tg^{-1}(\blue{x}) \] \[ \frac{1}{1 + \blue{x}^2} \] \[ cot^{-1}(\blue{x}) \] \[ -\frac{1}{1 + \blue{x}^2} \]
\[ sec^{-1}(\blue{x}) \] \[ \frac{1}{|\blue{x}|\sqrt{\blue{x}^2 - 1}} \] \[ csc^{-1}(\blue{x}) \] \[ -\frac{1}{|\blue{x}|\sqrt{\blue{x}^2 - 1}} \]

Intégrales

\[ \begin{aligned} & \int_\orange{a}^\orange{b} f(x)dx = \orange{-} \int_\orange{b}^\orange{a} f(x)dx \\[2em] & \int u' e^{u} dx = e^{u} + C \\[2em] & \int \frac{u'}{u} dx = ln(u) + C \\[2em] & \int \frac{u'}{1 + u^2} = tg^{-1} (u) + C \\[2em] & \int u'u^n dx = \frac{u^{n + 1}}{n + 1} + C \end{aligned} \]
Intégrales simples
Fonction Dérivée
\[ \frac{1}{3} \blue{u}^3 \] \[ \blue{u}^2 \blue{u}' \]
\[ \frac{1}{2} \blue{u}^2 \] \[ \blue{u} \blue{u}' \]
\[ \blue{u} \] \[ \blue{u}' \]
\[ ln(\blue{u}) \] \[ \frac{\blue{u}'}{\blue{u}} \]
\[ - \frac{1}{\blue{u}} \] \[ \frac{\blue{u}'}{\blue{u}^2} \]
\[- \frac{1}{2\blue{u}^2} \] \[ \frac{\blue{u}'}{\blue{u}^3} \]
Intégrales utiles
Fonction Dérivée
\[ \blue{x} ~ ln(\blue{x}) - x \] \[ ln(\blue{x}) \]
\[ ln(\blue{u}^n) \] \[ \frac{n}{\blue{u}} \]
\[ atg(\blue{u}) \] \[ \frac{\blue{u}'}{1 + \blue{u}^2} \]
\[ \frac{1}{a} atg(\frac{\blue{u}}{a}) \] \[ \frac{\blue{u}'}{a^2 + \blue{u}^2} \]
\[ (\blue{u} \blue{v})' - \blue{u}' \blue{v} \] \[ \blue{u} \blue{v}' \]

Taylor-Young et Maclaurin

\[ t(\blue{x}) = f(x_0) + f'(x_0)(\blue{x} - x_0) \] \[ \begin{aligned} & \text{linéarisation en } x_0 \\ & \text{fonction de la tangente en } x_0 \end{aligned}\]
\[ \begin{aligned} & T_1(\blue{x}) = f(x_0) + f'(x_0)(\blue{x} - x_0) + o(x) \\[2em] & T_2(\blue{x}) = f(x_0) + f'(x_0)(\blue{x} - x_0) + \frac{f''(x_0)}{2!}(\blue{x}-x_0) + o(x^2) \\[2em] & T_3(\blue{x}) = f(x_0) + f'(x_0)(\blue{x} - x_0) + \frac{f''(x_0)}{2!}(\blue{x}-x_0) + \frac{f'''(x_0)}{3!}(\blue{x}-x_0) +o(x^3) \end{aligned}\]
\[ T_n(\blue{x}) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!} (\blue{x} - x_0)^k + R_n(x) \] Formule générale
\[ \begin{aligned} \lim_{x \to x_0} \frac{f(x)}{g(x)} \quad = \quad \lim_{x \to x_0} \frac{F_n(x) + o(x^n)}{G_n(x) + o(x^n)} \quad \overset{x_0 = 0}{=} \quad \lim_{x \to x_0} \frac{f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^k(0)}{k!}x^k}{g(0) + g'(0)x + \frac{g''(0)}{2!}x^2 + \frac{g'''(0)}{3!}x^3 + ... + \frac{g^k(0)}{k!}x^k} \end{aligned}\]

Techniques

Intégration par parties

\[ (fg)' = f'g + fg' \quad\Longleftrightarrow\quad\quad \int fg' ~~=~~ fg ~~-~~ \int fg' \]
\[ \begin{aligned} & \int ln(x) ~dx \quad \begin{array}{l} & x && \overset{'}{\rightarrow} && 1 \\[1em] & ln(x) && \overset{'}{\rightarrow} && \frac{1}{x} \end{array} \\[3em] & \int 1 ~ ln(x) ~ dx \quad=\quad x ~ ln(x) ~~-~~ \int x \frac{1}{x} ~ dx \\[2em] & \int ~ ln(x) ~ dx \quad=\quad x ~ ln(x) ~~-~~ \int 1 ~ dx \\[2em] & \int ln(x) ~ dx \quad=\quad x ~ ln(x) ~~-~~ x \\[2em] \end{aligned} \]

Changement de variable

Rappel \[ f(x) = y \quad\Rightarrow\quad f'(x) = \frac{dy}{dx} \]
\[ \begin{aligned} & I = \int_{\green{0}}^{\green{1}} \sqrt{ 1 - \green{y}^2 } ~~ \green{dy} \\[2em] & \green{y} = cos(\orange{x}) \quad\Rightarrow\quad cos'(x) = \frac{\green{dy}}{\orange{dx}} = -sin(\orange{x}) \quad\Rightarrow\quad \green{dy} =-sin(\orange{x}) \orange{dx} \\[1em] & \green{y} = \green{0} \quad\Longleftrightarrow\quad \orange{x} = \orange{\frac{ \pi }{ 2 }} \\[1em] & \green{y} = \green{1} \quad\Longleftrightarrow\quad \orange{x} = \orange{0} \\[2em] & I = \int_{\orange{\pi / 2}}^{\orange{0}} \sqrt{1 - cos(\orange{x})^2 } ~~ (-sin(\orange{x})) \orange{dx} \gray{ \quad= - \int_{\pi / 2}^{0} \sqrt{sin(x)^2} ~ sin(x) ~ dx = \int_{0}^{\pi / 2} sin(x)^2 ~ dx } \\[2em] & = \underline{\underline{\frac{1}{4} \pi}} \end{aligned} \]
\[ \begin{aligned} & I = \int_{\orange{0}}^{\orange{\frac{\pi}{4}}} tg(\orange{x})^3 ~~ \orange{dx} \\[2em] & \green{y} = tg(\orange{x}) \quad\Rightarrow\quad tg'(\orange{x}) = \frac{\green{dy}}{\orange{dx}} = 1 + tg(\orange{x})^2 = 1 + \green{y}^2 \quad\Rightarrow\quad \orange{dx} = \frac{1}{1 + \green{y}^2} ~ \green{dy} \\[1em] & \green{y} = \green{0} \quad\Longleftrightarrow\quad \orange{x} = \orange{0} \gray{\quad\Longleftrightarrow y = tg(0)} \\[1em] & \green{y} = \green{1} \quad\Longleftrightarrow\quad \orange{x} = \orange{\frac{\pi}{4}} \gray{\quad\Longleftrightarrow y = tg(\frac{\pi}{4})} \\[2em] & I = \int_{\green{0}}^{\green{1}} \green{y}^3 \frac{1}{1 + \green{y}^2} ~ \green{dy} \gray{ \quad=\quad \int_{0}^{1} \frac{y^3}{1 +y^2 } ~~ dy \quad=\quad \int_{0}^{1} y - \frac{y}{1 +y^2 } ~~ dy \quad=\quad \int_{0}^{1} y ~ dy - \int_{0}^{1}\frac{y}{1 +y^2 } ~~ dy \quad=\quad \left[ \frac{1}{2} \right]_0^1 - \left[ \frac{1}{2} ln(1 + y^2)\right]_0^1 } \\[2em] & = \underline{\underline{ \frac{1}{2} - \frac{1}{2} ln(2) }} \end{aligned} \]