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junior college 1 | H3 Maths
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How do you derive this?
Date Posted: 3 years ago
Views: 702
By the chain rule, we get
d/dx [d2y/dx2]^2
= 2 [d2y/dx2]^1 times d/dx [d2y/dx2]
= 2 [d2y/dx2] [d3y/dx3]
d/dx [d2y/dx2]^2
= 2 [d2y/dx2]^1 times d/dx [d2y/dx2]
= 2 [d2y/dx2] [d3y/dx3]
See 1 Answer
done
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Hope this helps!
Date Posted:
3 years ago
Thanks
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