We have:
#x=e^sqrtt#
#y=1/t+2#
We manipulate the equations a bit.
#=>ln(x)=sqrtt#
#=>(ln(x))^2=t#
#=>y-2=1/t#
#=>1/(y-2)=t#
We now have:
#(ln(x))^2=1/(y-2)#
#=>1/(ln(x))^2=y-2#
#=>(ln(x))^-2+2=y#
#=>d/dx[(ln(x))^-2+2]=d/dx[y]#
#=>d/dx[(ln(x))^-2]+d/dx[2]=dy/dx#
Power rule:
#d/dx[x^n]=nx^(n-1)#
#d/dx[ln(x)]=1/x#
Chain rule:
#d/dx[f(g(x))]=f'(g(x))*g'(x)#
#=>-2(ln(x))^(-2-1)*d/dx[ln(x)]+0*2*x^(0-1)=dy/dx#
#=>-2(ln(x))^(-3)*1/x+0=dy/dx#
#=>-2/(ln(x))^(3)*1/x=dy/dx#
#=>-2/(x(ln(x))^(3))=dy/dx#
Substitute by using the fact that #e^sqrtt=x#
#=>-2/(e^sqrtt(ln(e^sqrtt))^(3))=dy/dx#
#=>-2/(e^sqrtt(t^(1/2))^(3))=dy/dx#
#=>-2/(e^sqrtt(t^(3/2)))=dy/dx#
Replace #t# with #1#.
#=>-2/(e^sqrt1(1^(3/2)))=f'(1)#
#=>-2/(e)=f'(1)#
