What is the derivative of y=2t(3t^2-5)^3 ?

The derivative of y=2t(3t^2-5)^3 is 2((3t^2-5)^3+18t^2(3t^2-5)^2)

What is the first derivative of y=2t(3t^2-5)^3 ?

The first derivative of y=2t(3t^2-5)^3 is 2((3t^2-5)^3+18t^2(3t^2-5)^2)

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derivative of

y = 2 t (3 t^{2} - 5)^{3}

2 ((3 t^{2} - 5)^{3} + 18 t^{2} (3 t^{2} - 5)^{2})

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = t, g = (3 t^{2} - 5)^{3}

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

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

\frac{d}{d t} ((3 t^{2} - 5)^{3}) = 18 t (3 t^{2} - 5)^{2}

= 2 (1 \cdot (3 t^{2} - 5)^{3} + 18 t (3 t^{2} - 5)^{2} t)

Simplify

2 (1 \cdot (3 t^{2} - 5)^{3} + 18 t (3 t^{2} - 5)^{2} t) : 2 ((3 t^{2} - 5)^{3} + 18 t^{2} (3 t^{2} - 5)^{2})

= 2 ((3 t^{2} - 5)^{3} + 18 t^{2} (3 t^{2} - 5)^{2})

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

\frac{d}{d x} (x 3 x - 2)

x y^{^{'}} + y = 0, y (4) = 6

d er i v a t i v e f (x) = \frac{4}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

t an g e n t y = - \frac{7}{2} x^{2} + 5 x + 5, (- 1, - \frac{7}{2})

What is the derivative of y=2t(3t^2-5)^3 ?
The derivative of y=2t(3t^2-5)^3 is 2((3t^2-5)^3+18t^2(3t^2-5)^2)
What is the first derivative of y=2t(3t^2-5)^3 ?
The first derivative of y=2t(3t^2-5)^3 is 2((3t^2-5)^3+18t^2(3t^2-5)^2)