"V=\\{a_{0}+a_{1}x+a_{2}x^{2}|a_{0},a_{1},a_{2}\\in R\\}\\\\\n\\\\\nW=<1+x^{2},1+2x^{2}>"
Any vector of W is of the form "a(1+x^{2})+b(1+2x^{2})" =(a+b)+(a+2b)"x^{2}"
Let T="\\dfrac{d}{dx}"
As T operates on W, according to the definition of differential operator
T("(a+b)+(a+2b)x^{2})=2x(a+2b)"
Hence Kernel (T)="\\{p\\in W|p=0\\}" where p is a polynomial of W.
"\\Rightarrow T ((a+b)+(a+2b)x^{2}=0"
"\\Rightarrow 2x(a+2b)=0"
"\\Rightarrow (a+2b)=0"
"\\Rightarrow a=-2b"
Therefore P=(-2b+b)+(-2b+2b)x"^{2}" =-b , which is a constant polynomial.
As "b\\in R\\\\" , Kernel(T) is isomorphic to R.
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