What is the d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) ?

The d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) is (3t^2)/4-4/(t^2)-sin(t-2)

What is the first d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) ?

The first d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) is (3t^2)/4-4/(t^2)-sin(t-2)

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\frac{d}{d t} ((\frac{t ^{3}}{4} - 2) + (\frac{4}{t} - 3) + cos (t - 2))

\frac{3 t ^{2}}{4} - \frac{4}{t ^{2}} - sin (t - 2)

Solution steps

\frac{d}{d t} ((\frac{t ^{3}}{4} - 2) + (\frac{4}{t} - 3) + cos (t - 2))

Simplify

(\frac{t ^{3}}{4} - 2) + (\frac{4}{t} - 3) + cos (t - 2) : \frac{t ^{3}}{4} + \frac{4}{t} + cos (t - 2) - 5

= \frac{d}{d t} (\frac{t ^{3}}{4} + \frac{4}{t} + cos (t - 2) - 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (\frac{t ^{3}}{4}) + \frac{d}{d t} (\frac{4}{t}) + \frac{d}{d t} (cos (t - 2)) - \frac{d}{d t} (5)

\frac{d}{d t} (\frac{t ^{3}}{4}) = \frac{3 t ^{2}}{4}

\frac{d}{d t} (\frac{4}{t}) = - \frac{4}{t ^{2}}

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

\frac{d}{d t} (5) = 0

= \frac{3 t ^{2}}{4} - \frac{4}{t ^{2}} - sin (t - 2) - 0

Simplify

= \frac{3 t ^{2}}{4} - \frac{4}{t ^{2}} - sin (t - 2)

\int \frac{- 4 x ^{3}}{2 x ^{3} + x ^{2} - 2 x - 1} d x

x \to 1 lim (5 x^{4} - 5 x^{2} + 10)

f^{^{'}} (x) = f (x)^{4} \cdot (4 x^{2} + x)

\int \frac{6 x - 7}{( 2 x + 1 ) ( 4 x ^{2} + 1 )} d x

s^{2} + 4 s + 13

What is the d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) ?
The d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) is (3t^2)/4-4/(t^2)-sin(t-2)
What is the first d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) ?
The first d/(dt)(((t^3)/4-2)+(4/t-3)+cos(t-2)) is (3t^2)/4-4/(t^2)-sin(t-2)