Hi.
If the distribution function of a distribution is:
f ( x ) = 1 − ( 1 − x a ) b {\displaystyle f(x)=1-(1-x^{a})^{b}}
Then: ∂ f ( x ) ∂ a = x a ln ( x ) b ( 1 − x a ) b − 1 {\displaystyle {\frac {\partial f(x)}{\partial a}}=x^{a}\ln(x)b(1-x^{a})^{b-1}}
∂ f ( x ) ∂ b = − ( 1 − x a ) b ln ( 1 − x a ) {\displaystyle {\frac {\partial f(x)}{\partial b}}=-(1-x^{a})^{b}\ln(1-x^{a})}
∂ f ( x ) ∂ x = a b ( 1 − x a ) b − 1 x a − 1 {\displaystyle {\frac {\partial f(x)}{\partial x}}=ab(1-x^{a})^{b-1}x^{a-1}}
If we take the logarithm of the likelihood function:-
ln ( f ( x ) ) = ln ( 1 − ( 1 − x a ) b ) {\displaystyle \ln(f(x))=\ln(1-(1-x^{a})^{b})}
∂ ln ( f ( x ) ) ∂ a = − x a b ( 1 − x a ) b − 1 ln ( x ) ( 1 − x a ) b − 1 = − x a b ln ( x ) {\displaystyle {\frac {\partial \ln(f(x))}{\partial a}}={\frac {-x^{a}b(1-x^{a})^{b-1}\ln(x)}{(1-x^{a})^{b-1}}}=-x^{a}b\ln(x)}
∂ ln ( f ( x ) ) ∂ b = ( 1 − x a ) b ln ( 1 − x a ) ( 1 − x a ) b − 1 = ( 1 − x a ) ln ( 1 − x a ) {\displaystyle {\frac {\partial \ln(f(x))}{\partial b}}={\frac {(1-x^{a})^{b}\ln(1-x^{a})}{(1-x^{a})^{b-1}}}=(1-x^{a})\ln(1-x^{a})}
∂ ln ( f ( x ) ) ∂ x = − a b ( 1 − x a ) b − 1 x a − 1 ( 1 − x a ) b − 1 = − a b x a − 1 {\displaystyle {\frac {\partial \ln(f(x))}{\partial x}}={\frac {-ab(1-x^{a})^{b-1}x^{a-1}}{(1-x^{a})^{b-1}}}=-abx^{a-1}}
