What is the d/(dt)((t^4-2t)^2-5) ?

The d/(dt)((t^4-2t)^2-5) is 2(4t^3-2)(t^4-2t)

What is the first d/(dt)((t^4-2t)^2-5) ?

The first d/(dt)((t^4-2t)^2-5) is 2(4t^3-2)(t^4-2t)

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\frac{d}{d t} ((t^{4} - 2 t)^{2} - 5)

2 (4 t^{3} - 2) (t^{4} - 2 t)

Solution steps

\frac{d}{d t} ((t^{4} - 2 t)^{2} - 5)

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= 2 (4 t^{3} - 2) (t^{4} - 2 t)

\int_{0}^{9} 81 - x^{2} d x

x^{y} = y^{x}

\int \frac{x ^{3}}{e ^{x^{4}}} d x

\frac{\partial}{\partial z} (3 y)

ln (1 - 2 x)

What is the d/(dt)((t^4-2t)^2-5) ?
The d/(dt)((t^4-2t)^2-5) is 2(4t^3-2)(t^4-2t)
What is the first d/(dt)((t^4-2t)^2-5) ?
The first d/(dt)((t^4-2t)^2-5) is 2(4t^3-2)(t^4-2t)