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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