Question 1027412
 {{{F(x) = (1+sin(pi*x))^h(x)}}}==>  {{{lnF(x) = h(x)ln(1+sin(pi*x))}}}

After differentiating this gives

{{{(dF(x)/dx)/F(x) = (dh(x)/dx)*ln(1+sin(pi*x)) + 
h(x)*((pi*cos(pi*x))/(1+sin(pi*x)))}}}

==> {{{(dF(1)/dx)/F(1) = (dh(1)/dx)*ln(1+sin(pi)) + 
h(1)*((pi*cos(pi))/(1+sin(pi)))}}}

But {{{F(1) = (1 + sin(pi))^h(1) = 1^2 = 1}}}

==> {{{dF(1)/dx = (dh(1)/dx)*ln1 + 
2*((-pi)/1) = -1*0 -2pi = -2pi}}}