What is the instantaneous velocity of an object moving in accordance to # f(t)= (e^(sqrtt),1/t+2) # at # t=1 #?

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Answer 1 · 2018-04-16T14:44:45

Please read below.

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#

#d/dx[x^n]=nx^(n-1)#

#d/dx[ln(x)]=1/x#

#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)#