Answer:
1) [tex]f'(t)=4t,\ g'(t)=3t^2+4[/tex]
2) [tex]p(t) =2t^5+8t^3[/tex]
[tex]p'(t)=10t^4+24t^2[/tex]
3) False
4)[tex]q(t) =\dfrac{1}{2}t+2t^{-1}[/tex]
[tex]q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}[/tex]
5) False
Step-by-step explanation:
Given that:
[tex]f(t) = 2t^2[/tex] and [tex]g(t) = t^3 + 4t[/tex]
Formula:
[tex]1. \dfrac{d}{dx}x^n=nx^{n-1}[/tex]
[tex]2. \dfrac{d}{dx}C.f(x)=C.f'(x)\ \{\text{C is a constant}\}[/tex]
1) Using above formula:
[tex]f'(t)=2\times 2 t^{2-1}=4t[/tex]
[tex]g'(t)=3t^{3-1}+4\times 1 t^{1-1}=3t^2+4[/tex]
2) [tex]p(t) =2t^2(t^3+4t)[/tex]
Rewriting the formula by distributing the [tex]2t^2[/tex] term:
[tex]p(t) =2t^2.t^3+2t^2.4t=2t^5+8t^3[/tex]
[tex]p'(t) = 10t^4+24t^2[/tex]
3) By using answers of part (1):
[tex]f'(t).g'(t)=12t^3+16t[/tex]
[tex]p'(t) = 10t^4+24t^2[/tex]
Therefore it is False that [tex]p'(t) = f'(t).g'(t)[/tex]
4) [tex]q(t)=\dfrac{t^3+4t}{2t^2}[/tex]
Writing by distributing:
[tex]q(t)=\dfrac{t^3}{2t^2}+\dfrac{4t}{2t^2}\\\Rightarrow q(t) =\dfrac{t}{2}+\dfrac{2}{t}\\\Rightarrow q(t) =\dfrac{1}{2}t+2t^{-1}[/tex]
Using the formula:
[tex]q'(t)=\dfrac{1}{2}t^{1-1}+2\dfrac{-1}{t^2}\\\Rightarrow q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}[/tex]
(5)By using answers in part (1):
[tex]\dfrac{g'(t)}{f'(t)}=\dfrac{3t^2+4}{4t}=\dfrac{3}{4}t+\dfrac{1}t[/tex]
[tex]q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}[/tex]
Therefore, it is False that:
[tex]q'(t)=\dfrac{g'(t)}{f'(t)}[/tex]
