How do i find the derivative?

Question

f(x)=t^4-5t

i don't think the h->0 approach will work seeing as how expanding (a+h)^4 is really hard, and i think it gets worse even after you factor.

do i use the x->a method? if so how do i go about this?

we aren't allowed to use power rule yet and must show work on exam/homework

Answer 1

Use the standard result:-
f (x)= a x ^n
f `(x) = na x^(n - 1)

f ( t ) = t^4 - 5t^1
f `( t ) = 4 t ³ - 5t^0
f `(t) = 4 t ³ - 5

Answer 2

the h->0 approach is to give u a working idea about derivative.
here is an easy approach.follow the steps:
1.if f(x)=t^n,then f'(x)=n*{t^(n-1)}.

2.f(x)=t^4-5t then f'(x)={4*t^(4-1)}-5*{1*t^(1-1)}

3. the last line evaluates to: f'(x)= 4t^3-5 (since t^0=1)

Answer 3

Use power rule:
f'(t) = 4t^3 - 5

Answer 4

just differientaite it.

f ' (x) = 4t^3 - 5

Answer 5

(a+h)^4 = a^4+4a^3*h+6a^2h^2+4ah^3+h^4

Does that help?

So -- [f(a+h)-f(a)]/h =

[(a^4+4a^3*h+6a^2h^2+4ah^3+h^4 -5a-5h)-a^4-5a]/h =

[4a^3*h+6a^2h^2+4ah^3+h^4-5h]/h =

4a^3+6a^2h+4ah^2+h^3-5

Take the limit as h==0 and you get

4a^3-5

Answer 6

id say stick it out with first principles. it may get ugly but do'er, cause obviously you havent learned the power rule