We have: #(sinx)^(1/x^2)#
We rewrite this as:
#(sinx)^(x^-2)#
Remember this rule:
#d/dx[f(x)^(g(x))]=f(x)^(g(x))*d/dx[ln(f(x))*g(x)]#
Also, remember the following:
#d/dx[lnx]=1/x#
#d/dx[f(x)*g(x)]=f'(x)*g(x)+f(x)*g'(x)#
#d/dx[f(g(x))]=f'(g(x))*g'(x)#
We now have:
#(sinx)^(x^-2)*d/dx[ln(sinx)*x^-2]#
#(sinx)^(x^-2)*(d/dx[ln(sinx)]*x^-2+ln(sinx)*d/dx[x^-2])#
Some other rules:
#d/dx[x^n]=nx^(n-1)#
#d/dx[sinx]=cosx#
#=>(sinx)^(x^-2)*(1/(sinx)*x^-2*d/dx[sinx]+ln(sinx)(-2x^-3))#
#=>(sinx)^(x^-2)*(1/(sinx)*x^-2*cosx+ln(sinx)(-2x^-3))#
#=>(sinx)^(1/x^2)*(1/(sinx)*cosx*1/x^2+ln(sinx)(-2x^-3))#
#=>(sinx)^(1/x^2)*(cosx/((sinx)x^2)+ln(sinx)(-2x^-3))#
#=>(sinx)^(1/x^2)*(cosx/((sinx)x^2)-2ln(sinx)(1/x^3))#
#=>(sinx)^(1/x^2)*(cosx/(x^2sinx)-(2ln(sinx))/x^3)#
This is our answer!