Explications sur la dérivation 5

Explications sur la dérivation 5𝑥² − 8𝑥 + 3

(𝑢 + 𝑣)^′= 𝑢^′+ 𝑣′

(5𝑥²− 8𝑥 + 3)^′ = (5𝑥²)^′− (8𝑥)^′+ (3)^′

(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 × 𝑓𝑜𝑛𝑐𝑡𝑖𝑜𝑛)^′= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 × (𝑓𝑜𝑛𝑐𝑡𝑖𝑜𝑛)′

(𝑘 × 𝑢(𝑥))^′ = 𝑘 × 𝑢′(𝑥)

(5𝑥²− 8𝑥 + 3)^′ = (5𝑥²)^′− (8𝑥)^′+ (3)^′

= 5 × (𝑥²)^′− 8 × (𝑥)^′+ (3)^′ (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒)^′= 0

(𝑥)^′= 1 (𝑥²)^′= 2𝑥

…

(𝑥^𝑛)^′= 𝑛 × 𝑥^𝑛−1 𝑒𝑥 ∶ (𝑥³)^′= 3𝑥² (𝑥²⁰²¹)^′= 2021𝑥²⁰²⁰

(𝟓𝒙^𝟐− 𝟖𝒙 + 𝟑)^′= (5𝑥²)^′− (8𝑥)^′+ (3)^′

= 5 × (𝑥²)^′− 8 × (𝑥)^′+ (3)^′

= 5 × 2𝑥 − 8 × 1 + 0

= 𝟏𝟎𝒙 − 𝟖

𝟓𝒙 + 𝟏

𝟐𝒙²+ 𝟑 𝒆𝒔𝒕 𝒅𝒖 𝒕𝒚𝒑𝒆𝒖(𝒙) 𝒗(𝒙) (𝑢

𝑣)

′

=𝑢^′× 𝑣 − 𝑢 × 𝑣^′ 𝑣²

𝑖𝑐𝑖 𝑢 = 5𝑥 + 1 𝑒𝑡 𝑣 = 2𝑥²+ 3 𝑑𝑜𝑛𝑐 𝑢^′ = (5𝑥)^′+ (1)^′= 5 + 0 = 5

𝑒𝑡 𝑣^′= (2𝑥²)^′+ (3)^′= 2 × (𝑥²)^′+ 0 = 2 × 2𝑥 + 0 = 4𝑥 𝑢^′× 𝑣 − 𝑢 × 𝑣^′

𝑣² =5 × (2𝑥²+ 3) − (5𝑥 + 1) × 4𝑥 (2𝑥²+ 3)²

=10𝑥²+ 15 − 20𝑥²− 4𝑥 (2𝑥²+ 3)²

=−𝟏𝟎𝒙^𝟐− 𝟒𝒙 + 𝟏𝟓 (𝟐𝒙^𝟐+ 𝟑)²

−𝟑𝒆^𝒙 𝒆𝒔𝒕 𝒅𝒖 𝒕𝒚𝒑𝒆 𝒌 × 𝒖(𝒙) 𝑎𝑣𝑒𝑐 𝑘 = −3 𝑒𝑡 𝑢(𝑥) = 𝑒^𝑥

(𝑘 × 𝑢(𝑥))^′= 𝑘 × 𝑢^′(𝑥)

(−𝟑𝒆^𝒙)^′= −3 × (𝑒^𝑥)^′= −3 × 𝑒^𝑥 = −𝟑𝒆^𝒙