What is the derivative of y=(t^2+8)^2-2(t^2+8) ?

The derivative of y=(t^2+8)^2-2(t^2+8) is 4t^3+28t

What is the first derivative of y=(t^2+8)^2-2(t^2+8) ?

The first derivative of y=(t^2+8)^2-2(t^2+8) is 4t^3+28t

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derivative of

y = (t^{2} + 8)^{2} - 2 (t^{2} + 8)

4 t^{3} + 28 t

Solution steps

\frac{d}{d t} ((t^{2} + 8)^{2} - 2 (t^{2} + 8))

Simplify

(t^{2} + 8)^{2} - 2 (t^{2} + 8) : t^{4} + 14 t^{2} + 48

= \frac{d}{d t} (t^{4} + 14 t^{2} + 48)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (t^{4}) + \frac{d}{d t} (14 t^{2}) + \frac{d}{d t} (48)

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

\frac{d}{d t} (14 t^{2}) = 28 t

\frac{d}{d t} (48) = 0

= 4 t^{3} + 28 t + 0

Simplify

= 4 t^{3} + 28 t

\frac{\partial}{\partial z} (e^{- x^{2} - 2 y^{2} - 3 z^{2}})

\int (3 x + 1)^{11} d x

\int_{0}^{6} x e^{- x} d x

\frac{d}{d x} (\frac{( 5 x ^{4} + x ^{2} )}{2})

What is the derivative of y=(t^2+8)^2-2(t^2+8) ?
The derivative of y=(t^2+8)^2-2(t^2+8) is 4t^3+28t
What is the first derivative of y=(t^2+8)^2-2(t^2+8) ?
The first derivative of y=(t^2+8)^2-2(t^2+8) is 4t^3+28t