Asked by Syam

Find second differentiation for function x^3=ycosx
Answered by Reiny
I assume you differentiating with respect to x ?

3x^2 = y(-sinx) + dy/dx(cosx)
dy/dx = 3x^2 + ysinx
d(dy/dx)/dx = 6x + dy/dx sinx + ycosx
= 6x + (3x^2 + ysinx)sinx + ycosx
= 6x^2 + 3x^2sinx + ysin^2 x + ycosx

check my work, should have written it on paper first
Answered by Reiny
or

y = x^3/cosx
y' = (cosx(3x^2) - x^3(-sinx))/cos^2x
= (3x^2 cosx + x^3 sinx)/cos^2x
now do it again
Answered by Syam
Why 6x become power of 2 at the end?
Answered by Reiny
That was clearly a typo, good for you to catch it
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