Differentiate `sinx^(2)+sin^(2)x+sin^(2)(x^(2))`

Explanation

First of all we make the coefficient of x as 1 in each factor by taking some constant common as :
`2(x+(7)/(2))(x+3)(x-3).2(x-(5)/(2))+9`
Now, we take the product of 2 factors at a time in such a way that sum of constant terms is same in each pair as :
` (##NTN_MATH_IX_C02_S01_066_S01.png" width="80%">
`=4(x^(2)+(x)/(2)-(21)/(2))(x^(2)+(x)/(2)-(15)/(2))+9`
Let `x^(2)+(x)/(2)=a`
`=(a-(21)/(2))(a-(15)/(2))+9`
`4(a^(2)-18a+(315)/(4))+9`
`4(a^(2)-18a)+315+9`
`4underset("make perfect square")(ubrace(a^(2)-18a+81)-81)+324 " " ["add and subtract"(("coeff. of a")/(2))^(2) " to make perfect square"]`
`=4(a^(2)-18a+81)-324+324`
`=4(a-9)^(2)`
Replace the value of a as `x^(2)+(x)/(2)`, we get
`4(x^(2)+(x)/(2)-9)^(2)=4((2x^(2)+x-18)/(2))^(2)=(2x^(2)+x-18)^(2)`
`therefore " Square root of" (2x+7)(x^(2)-9)(2x-5)+9 i.e., (2x^(2)+x-18)^(2) " is" 2x^(2)+x-18`.