What is the derivative of sqrt(16+t) ?

The derivative of sqrt(16+t) is 1/(2sqrt(16+t))

What is the first derivative of sqrt(16+t) ?

The first derivative of sqrt(16+t) is 1/(2sqrt(16+t))

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derivative of

16 + t

\frac{1}{2 16 + t}

Solution steps

\frac{d}{d t} (16 + t)

Apply the chain rule:

\frac{1}{2 16 + t} \frac{d}{d t} (16 + t)

= \frac{1}{2 16 + t} \frac{d}{d t} (16 + t)

\frac{d}{d t} (16 + t) = 1

= \frac{1}{2 16 + t} \cdot 1

Simplify

= \frac{1}{2 16 + t}

\frac{d}{d x} ((5 x + 2)^{4} (3 x - 7))

\int e^{- x^{2}} \cdot x d x

k = 1 \sum \infty \frac{k ^{2}}{k !}

\frac{d}{d x} (\frac{x ^{3} + 2 x ^{2} - 1}{x + 5})

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What is the derivative of sqrt(16+t) ?
The derivative of sqrt(16+t) is 1/(2sqrt(16+t))
What is the first derivative of sqrt(16+t) ?
The first derivative of sqrt(16+t) is 1/(2sqrt(16+t))