What is the derivative of (t+3)(2t-1)^2 ?

The derivative of (t+3)(2t-1)^2 is 12t^2+16t-11

What is the first derivative of (t+3)(2t-1)^2 ?

The first derivative of (t+3)(2t-1)^2 is 12t^2+16t-11

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derivative of

(t + 3) (2 t - 1)^{2}

12 t^{2} + 16 t - 11

Solution steps

\frac{d}{d t} ((t + 3) (2 t - 1)^{2})

Apply the Product Rule:

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

f = t + 3, g = (2 t - 1)^{2}

= \frac{d}{d t} (t + 3) (2 t - 1)^{2} + \frac{d}{d t} ((2 t - 1)^{2}) (t + 3)

\frac{d}{d t} (t + 3) = 1

\frac{d}{d t} ((2 t - 1)^{2}) = 4 (2 t - 1)

= 1 \cdot (2 t - 1)^{2} + 4 (2 t - 1) (t + 3)

Simplify

1 \cdot (2 t - 1)^{2} + 4 (2 t - 1) (t + 3) : 12 t^{2} + 16 t - 11

= 12 t^{2} + 16 t - 11

n = 0 \sum \infty n \cdot (\frac{4}{5})^{n}

y = \frac{529 - 8 x}{7 + 7 x}, (4, 7)

\frac{d}{d x} (\frac{a x ^{2} + b x + 1}{x ^{2} - 6 x + 8})

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

f (x) = x^{2} + 1, at x = 1

What is the derivative of (t+3)(2t-1)^2 ?
The derivative of (t+3)(2t-1)^2 is 12t^2+16t-11
What is the first derivative of (t+3)(2t-1)^2 ?
The first derivative of (t+3)(2t-1)^2 is 12t^2+16t-11