What is the derivative of sqrt(18t^2+9) ?

The derivative of sqrt(18t^2+9) is (6t)/(sqrt(2t^2+1))

What is the first derivative of sqrt(18t^2+9) ?

The first derivative of sqrt(18t^2+9) is (6t)/(sqrt(2t^2+1))

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derivative of

18 t^{2} + 9

\frac{6 t}{2 t ^{2} + 1}

Solution steps

\frac{d}{d t} (18 t^{2} + 9)

Apply the chain rule:

\frac{1}{2 18 t ^{2} + 9} \frac{d}{d t} (18 t^{2} + 9)

= \frac{1}{2 18 t ^{2} + 9} \frac{d}{d t} (18 t^{2} + 9)

\frac{d}{d t} (18 t^{2} + 9) = 36 t

= \frac{1}{2 18 t ^{2} + 9} \cdot 36 t

Simplify

\frac{1}{2 18 t ^{2} + 9} \cdot 36 t : \frac{6 t}{2 t ^{2} + 1}

= \frac{6 t}{2 t ^{2} + 1}

What is the derivative of sqrt(18t^2+9) ?
The derivative of sqrt(18t^2+9) is (6t)/(sqrt(2t^2+1))
What is the first derivative of sqrt(18t^2+9) ?
The first derivative of sqrt(18t^2+9) is (6t)/(sqrt(2t^2+1))