If y = 3 cot x/8, the double differentiation of y gives,
y'' = (3/32) (cosec x/8)(cot x/8)
Given value of y is,
y = 3 cot x/8
Differentiation of the above equation will give us the following,
y' = d(3 cot x/8) / dx
y' = 3d(cot x/8) / dx ........... (1)
Now, we know that the differentiation of cot x is -cosec²x
Therefore, d(cot x/8) / dx = -(cosec²x/8)/8
Thus, equation (1) transforms as follows,
y' = -3(cosec²x/8)/8
Taking differentiation of the above equation, we get,
y'' = -3d((cosec²x/8)/ 8dx
y'' = -6d(cosec x/8) / 8dx [Using chain rule] .......... (2)
As we know, the differentiation of cosec x is -(cosec x)(cot x),
d(cosec x/8) / dx = -(cosec x/8)(cot x/8) / 8 [Using chain rule]
Therefore, equation (2) can be written as,
y'' = -(-6(cosec x/8)(cot x/8) / (8×8)
∴ y '' = 3(cosec x/8)(cot x/8) / 32
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