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

1 Answer
Apr 16, 2018

Please read below.

Explanation:

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)