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

The derivative of (1+3t)^2 is 6(1+3t)

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

The first derivative of (1+3t)^2 is 6(1+3t)

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derivative of

(1 + 3 t)^{2}

6 (1 + 3 t)

Solution steps

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

Apply the chain rule:

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

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

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

= 2 (1 + 3 t) \cdot 3

Simplify

= 6 (1 + 3 t)

\frac{d}{d t} (t sin (t) + cos (t))

\int \frac{( 1 - x )}{x ^{2}} d x

(1 + x)^{4}

\int \frac{1}{sin ( x )} d x

\frac{1 - 2 t}{1 + 3 t}

What is the derivative of (1+3t)^2 ?
The derivative of (1+3t)^2 is 6(1+3t)
What is the first derivative of (1+3t)^2 ?
The first derivative of (1+3t)^2 is 6(1+3t)