What is the derivative of (2t+4)i+sin(4t)j+2t^3k ?

The derivative of (2t+4)i+sin(4t)j+2t^3k is (4jcos(4t)+6kt^2)+2i

What is the first derivative of (2t+4)i+sin(4t)j+2t^3k ?

The first derivative of (2t+4)i+sin(4t)j+2t^3k is (4jcos(4t)+6kt^2)+2i

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derivative of

(2 t + 4) i + sin (4 t) j + 2 t^{3} k

(4 j cos (4 t) + 6 k t^{2}) + 2 i

Solution steps

\frac{d}{d t} ((2 t + 4) i + sin (4 t) j + 2 t^{3} k)

Simplify

(2 t + 4) i + sin (4 t) j + 2 t^{3} k : (j sin (4 t) + 2 k t^{3}) + i (4 + 2 t)

= \frac{d}{d t} ((j sin (4 t) + 2 k t^{3}) + i (4 + 2 t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (j sin (4 t) + 2 k t^{3}) + \frac{d}{d t} (i (4 + 2 t))

\frac{d}{d t} (j sin (4 t) + 2 k t^{3}) = j cos (4 t) \cdot 4 + 6 k t^{2}

\frac{d}{d t} (i (4 + 2 t)) = 2 i

= j cos (4 t) \cdot 4 + 6 k t^{2} + 2 i

Group the real part and the imaginary part of the complex number

= (4 j cos (4 t) + 6 k t^{2}) + 2 i

x \to 0 + lim (x^{3 x})

y = \frac{2}{x}

y = \frac{13}{ln ( x )}

\int sin (t) cos^{2} (t) d t

x \to 0 lim (x^{3} + 2 x^{2} - 4 x + 4)

What is the derivative of (2t+4)i+sin(4t)j+2t^3k ?
The derivative of (2t+4)i+sin(4t)j+2t^3k is (4jcos(4t)+6kt^2)+2i
What is the first derivative of (2t+4)i+sin(4t)j+2t^3k ?
The first derivative of (2t+4)i+sin(4t)j+2t^3k is (4jcos(4t)+6kt^2)+2i