What is the derivative of 5(t)i+(40-5(t)^2)j ?

The derivative of 5(t)i+(40-5(t)^2)j is -10jt+5i

What is the first derivative of 5(t)i+(40-5(t)^2)j ?

The first derivative of 5(t)i+(40-5(t)^2)j is -10jt+5i

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derivative of

5 (t) i + (40 - 5 (t)^{2}) j

- 10 j t + 5 i

Solution steps

\frac{d}{d t} (5 t i + (40 - 5 t^{2}) j)

Simplify

5 t i + (40 - 5 t^{2}) j : (40 j - 5 j t^{2}) + 5 i t

= \frac{d}{d t} ((40 j - 5 j t^{2}) + 5 i t)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (40 j - 5 j t^{2}) + \frac{d}{d t} (5 i t)

\frac{d}{d t} (40 j - 5 j t^{2}) = - 10 j t

\frac{d}{d t} (5 i t) = 5 i

= - 10 j t + 5 i

What is the derivative of 5(t)i+(40-5(t)^2)j ?
The derivative of 5(t)i+(40-5(t)^2)j is -10jt+5i
What is the first derivative of 5(t)i+(40-5(t)^2)j ?
The first derivative of 5(t)i+(40-5(t)^2)j is -10jt+5i