Calculatrice de dérivées partielles
Calculer les dérivées partielles étape par étape
Cette calculatrice en ligne calculera la dérivée partielle de la fonction, avec les étapes indiquées. Vous pouvez spécifier n'importe quel ordre d'intégration.
Solution
Your input: find ∂2∂y2(2x2y−2x2+y3−2y2+2)
First, find ∂∂y(2x2y−2x2+y3−2y2+2)
The derivative of a sum/difference is the sum/difference of derivatives:
∂∂y(2x2y−2x2+y3−2y2+2)=(∂∂y(2)−∂∂y(2x2)−∂∂y(2y2)+∂∂y(y3)+∂∂y(2x2y))Apply the constant multiple rule ∂∂y(c⋅f)=c⋅∂∂y(f) with c=2x2 and f=y:
∂∂y(2x2y)+∂∂y(2)−∂∂y(2x2)−∂∂y(2y2)+∂∂y(y3)=2x2∂∂y(y)+∂∂y(2)−∂∂y(2x2)−∂∂y(2y2)+∂∂y(y3)Apply the power rule ∂∂y(yn)=n⋅y−1+n with n=1, in other words ∂∂y(y)=1:
2x2∂∂y(y)+∂∂y(2)−∂∂y(2x2)−∂∂y(2y2)+∂∂y(y3)=2x21+∂∂y(2)−∂∂y(2x2)−∂∂y(2y2)+∂∂y(y3)The derivative of a constant is 0:
2x2−∂∂y(2x2)+∂∂y(2)−∂∂y(2y2)+∂∂y(y3)=2x2−(0)+∂∂y(2)−∂∂y(2y2)+∂∂y(y3)Apply the constant multiple rule ∂∂y(c⋅f)=c⋅∂∂y(f) with c=2 and f=y2:
2x2−∂∂y(2y2)+∂∂y(2)+∂∂y(y3)=2x2−(2∂∂y(y2))+∂∂y(2)+∂∂y(y3)Apply the power rule ∂∂y(yn)=n⋅y−1+n with n=2:
2x2−2∂∂y(y2)+∂∂y(2)+∂∂y(y3)=2x2−2(2y−1+2)+∂∂y(2)+∂∂y(y3)=2x2−4y+∂∂y(2)+∂∂y(y3)The derivative of a constant is 0:
2x2−4y+∂∂y(2)+∂∂y(y3)=2x2−4y+(0)+∂∂y(y3)Apply the power rule ∂∂y(yn)=n⋅y−1+n with n=3:
2x2−4y+∂∂y(y3)=2x2−4y+(3y−1+3)=2x2+3y2−4yThus, ∂∂y(2x2y−2x2+y3−2y2+2)=2x2+3y2−4y
Next, ∂2∂y2(2x2y−2x2+y3−2y2+2)=∂∂y(∂∂y(2x2y−2x2+y3−2y2+2))=∂∂y(2x2+3y2−4y)
The derivative of a sum/difference is the sum/difference of derivatives:
∂∂y(2x2+3y2−4y)=(∂∂y(2x2)−∂∂y(4y)+∂∂y(3y2))Apply the constant multiple rule ∂∂y(c⋅f)=c⋅∂∂y(f) with c=3 and f=y2:
∂∂y(3y2)+∂∂y(2x2)−∂∂y(4y)=(3∂∂y(y2))+∂∂y(2x2)−∂∂y(4y)Apply the power rule ∂∂y(yn)=n⋅y−1+n with n=2:
3∂∂y(y2)+∂∂y(2x2)−∂∂y(4y)=3(2y−1+2)+∂∂y(2x2)−∂∂y(4y)=6y+∂∂y(2x2)−∂∂y(4y)Apply the constant multiple rule ∂∂y(c⋅f)=c⋅∂∂y(f) with c=4 and f=y:
6y−∂∂y(4y)+∂∂y(2x2)=6y−(4∂∂y(y))+∂∂y(2x2)Apply the power rule ∂∂y(yn)=n⋅y−1+n with n=1, in other words ∂∂y(y)=1:
6y−4∂∂y(y)+∂∂y(2x2)=6y−41+∂∂y(2x2)The derivative of a constant is 0:
6y−4+∂∂y(2x2)=6y−4+(0)Thus, ∂∂y(2x2+3y2−4y)=6y−4
Therefore, ∂2∂y2(2x2y−2x2+y3−2y2+2)=6y−4
Answer: ∂2∂y2(2x2y−2x2+y3−2y2+2)=6y−4