"Find the first and second derivatives of the..." solved and explained

Ben Shaver

Answered question

2021-12-15

Find the first and second derivatives of the function.

g (t) = 2 \cos t - 3 \sin

Bob Huerta

Beginner2021-12-16Added 41 answers

Step 1
We find the first derivative

g (t) = 2 \cos t - 3 \sin

g^{'} (t) = 2 (- \sin t) - 3 \cos

g^{'} (t) = - 2 \sin t - 3 \cos

Step 2
Then we find the second derivative

g^{'} (t) = - 2 \sin t - 3 \cos

g^{″} (t) = - 2 \cos t - 3 (- \sin t)

g^{″} (t) = - 2 \cos t + 3 \sin

Answer: First derivative:

g^{'} (t) = - 2 \sin t - 3 \cos

Second derivative:

g^{″} (t) = - 2 \cos t + 3 \sin

esfloravaou

Beginner2021-12-17Added 43 answers

Step 1
Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (2 \cos (t)) - \frac{d}{d t} (3 \sin (t))

\frac{d}{d t} (2 \cos (t)) = - 2 \sin (t)

\frac{d}{d t} (3 \sin (t)) = 3 \cos (t)

= - 2 \sin (t) - 3 \cos (t)

Step 2

\frac{d^{2}}{{d t}^{2}} (2 \cos (t) - 3 \sin (t))

\frac{d}{d t} (2 \cos (t) - 3 \sin (t)) = - 2 \sin (t) - 3 \cos (t)

= \frac{d}{d t} (- 2 \sin (t) - 3 \cos (t))

\frac{d}{d t} (- 2 \sin (t) - 3 \cos (t)) = - 2 \cos (t) + 3 \sin (t)

= - 2 \cos (t) + 3 \sin (t)

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