We know that h is a vector-valued function. This means it takes one number as an input (t) ,but it outputs two numbers as a two-dimensional vector.
Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as [tex]u(t)=(v(t),w(t))[/tex] then its derivative is the vector-valued function [tex] u'(t)=(v'(t),w'(t))[/tex] .
In other words, the derivative is found by differentiating each of the expressions in the function's output vector.
Recall that [tex]h(t)=(2t^3+6,e^{2t})[/tex]
Let's differentiate the first expression:
[tex]\dfrac{d}{dt} (2t^3+6)=3*2t^2) = 6t^2[/tex]
Let's differentiate the second expression:
[tex]\dfrac{d}{dt}(e^2^t)= 2e^2^t[/tex]
Now let's put everything together:
[tex]h′(t) = \dfrac{d}{dt}(2t^3+6),\dfrac{d}{dt}(e^2^t))
[/tex]
[tex]= 6t^2,2e^2^t[/tex]
In conclusion, [tex]h'(t) = (6t^2,2e^2^t)[/tex]
