What is the Derivative of Square Root of x^2 + 1?

The Mathematician — Tue, 22 Nov 2022 13:00:53 +0000

The derivative of

\sqrt{x^2 + 1}

\frac{x}{\sqrt{x^2 + 1}}

.

Solution. Let

F(x) = \sqrt{x^2 + 1}

f(u) = \sqrt{u} = u^{\frac{1}{2}}

and

g(x) = x^2 + 1

such that

F(x) = f(g(x))

. Using the chain rule, we can find the derivative of

\sqrt{x^2 + 1}

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We know from here that

f'(u) = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}

and that

g'(x) = 2x

. So we get:

\begin{align*} f'(g(x)) = \frac{1}{2\sqrt{x^2 + 1}}. \end{align*}

Substituting everything, we get:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x \\ &= \frac{x}{\sqrt{x^2 + 1}}. \end{align*}

So, the derivative of

\sqrt{x^2 + 1}

\frac{x}{\sqrt{x^2 + 1}}

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