What is the derivative of y=sqrt(x+1) ?

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

What is the first derivative of y=sqrt(x+1) ?

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

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derivative of

y = x + 1

\frac{1}{2 x + 1}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

f (x) = (x - 1) \cdot ln (x - 1)

\frac{d}{d x} (- e^{x - 1})

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

\frac{d}{d x} (e^{\frac{2 x}{2}})

\int \frac{cos ( 2 x )}{2} d x

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