Why derivative of d(2at)/dt is 2a?

Question

I was studying about parametric differentiation and in the book I saw that the derivative of the following below is 2a and I don't understand why implicit differentiation isn't applied here..
#d/dt (2at)# = 2a * 1 = 2a

Isn't it supposed to be:
#d/dt(2at)#
=#2 ( d/dt (a) * t + d/dt (t) * a )#
=#2 ( t (da)/dt+ a )#

1500 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-11T13:51:26

Because #a# is a parameter, not a function of #t#.

Using the limit definition of derivative, if #a# is an indeterminate quantity but is not dependent on #t# we have:

#d/dt (2at) = lim_(h->0) ( 2a(t+h) -2at)/h#

#d/dt (2at) = lim_(h->0) ( 2at+2ah -2at)/h#

#d/dt (2at) = lim_(h->0) ( 2ah )/h#

#d/dt (2at) = lim_(h->0) 2a = 2a#

On the other hand if #a=a(t)# is a function of #t# then we have:

#d/dt (2a(t)t) = lim_(h->0) ( 2a(t+h)(t+h) -2a(t)t)/h#

#d/dt (2a(t)t) = lim_(h->0) ( 2a(t+h)t+2a(t+h)h -2a(t)t)/h#

In this case since in general #a(t+h) != a(t)# it is not possible to simplify, but we have to proceed differently:

#d/dt (2a(t)t) = lim_(h->0) ( 2t (a(t+h)-a(t)))/h+2a(t+h)#

#d/dt (2a(t)t) = 2t lim_(h->0) (a(t+h)-a(t))/h+ lim_(h->0) 2a(t+h)#

and if #a(t)# is differentiable, so that:

#lim_(h->0) (a(t+h)-a(t))/h = a'(t)#

then:

#d/dt (2a(t)t) = 2ta'(t) + 2a(t)#

which is the result of the product rule.