What is the derivative of-3t(6t^3-1)^8 ?

The derivative of-3t(6t^3-1)^8 is -3((6t^3-1)^8+144t^3(6t^3-1)^7)

What is the first derivative of-3t(6t^3-1)^8 ?

The first derivative of-3t(6t^3-1)^8 is -3((6t^3-1)^8+144t^3(6t^3-1)^7)

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derivative of

- 3 t (6 t^{3} - 1)^{8}

- 3 ((6 t^{3} - 1)^{8} + 144 t^{3} (6 t^{3} - 1)^{7})

Solution steps

\frac{d}{d t} (- 3 t (6 t^{3} - 1)^{8})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 3 \frac{d}{d t} (t (6 t^{3} - 1)^{8})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = (6 t^{3} - 1)^{8}

= - 3 (\frac{d t}{d t} (6 t^{3} - 1)^{8} + \frac{d}{d t} ((6 t^{3} - 1)^{8}) t)

\frac{d t}{d t} = 1

\frac{d}{d t} ((6 t^{3} - 1)^{8}) = 144 t^{2} (6 t^{3} - 1)^{7}

= - 3 (1 \cdot (6 t^{3} - 1)^{8} + 144 t^{2} (6 t^{3} - 1)^{7} t)

Simplify

- 3 (1 \cdot (6 t^{3} - 1)^{8} + 144 t^{2} (6 t^{3} - 1)^{7} t) : - 3 ((6 t^{3} - 1)^{8} + 144 t^{3} (6 t^{3} - 1)^{7})

= - 3 ((6 t^{3} - 1)^{8} + 144 t^{3} (6 t^{3} - 1)^{7})

\int_{0}^{π} x sin (x) d x

\frac{d}{d x} ((- 2 x^{2} - 2 x + 3)^{5})

\int cos (x) sin (x) cos^{2} (x) + 1 d x

\int_{0}^{x} e^{x^{2}} x d x

\frac{\partial}{\partial x} (\frac{8 y}{x ^{5}})

What is the derivative of-3t(6t^3-1)^8 ?
The derivative of-3t(6t^3-1)^8 is -3((6t^3-1)^8+144t^3(6t^3-1)^7)
What is the first derivative of-3t(6t^3-1)^8 ?
The first derivative of-3t(6t^3-1)^8 is -3((6t^3-1)^8+144t^3(6t^3-1)^7)