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

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

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

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

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

\frac{1}{2 16 + x}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

x \to 0 lim (2^{- x})

\int_{1}^{2} x - 1 - x + 1 d x

f (x) = x^{s i n (x)}, (\frac{π}{2}, \frac{π}{2})

\int x^{2} (x^{3} - 1)^{3} d x

y = x, y = (x - 2)^{2}

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