What is the derivative of #f(t) = (e^(t^2-1)-e^t, 2t^2-4t ) #?

Question

1642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-08T07:53:36

Knowing the chain rule for derivatives, you can see that

#dx/dt=2te^(t^2-1)-e^(t)#

and

#dy/dt=4t-4#

Thus, #f'(t)=(2te^(t^2-1)-e^(t), 4t-4)#

In this case, the "trick" is to apply the chain rule to the the first term of the expression

#e^(t^2-1)-e^t#

to obtain the derivative.

The chain rule formula used in this problem comes from

#d/dx[e^(z)]=e^(z) dz/dx#

where

#z=t^2-1#