Find the Derivative – d/dx y=e^( square root of x)

$y = e^{\sqrt{x}}$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$\frac{d}{d x} [e^{x^{\frac{1}{2}}}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x^{\frac{1}{2}}$ .

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To apply the Chain Rule, set $u$ as $x^{\frac{1}{2}}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{\frac{1}{2}}]$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$e^{u} \frac{d}{d x} [x^{\frac{1}{2}}]$

Replace all occurrences of $u$ with $x^{\frac{1}{2}}$ .

$e^{x^{\frac{1}{2}}} \frac{d}{d x} [x^{\frac{1}{2}}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = \frac{1}{2}$ .

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{\frac{1}{2} - 1})$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{\frac{1}{2} - 1 \cdot \frac{2}{2}})$

Combine $- 1$ and $\frac{2}{2}$ .

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}})$

Combine the numerators over the common denominator.

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{\frac{1 - 1 \cdot 2}{2}})$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{\frac{1 - 2}{2}})$

Subtract $2$ from $1$ .

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{\frac{- 1}{2}})$

Move the negative in front of the fraction.

$e^{x^{\frac{1}{2}}} (\frac{1}{2} x^{- \frac{1}{2}})$

Combine $\frac{1}{2}$ and $x^{- \frac{1}{2}}$ .

$e^{x^{\frac{1}{2}}} \frac{x^{- \frac{1}{2}}}{2}$

Combine $e^{x^{\frac{1}{2}}}$ and $\frac{x^{- \frac{1}{2}}}{2}$ .

$\frac{e^{x^{\frac{1}{2}}} x^{- \frac{1}{2}}}{2}$

Move $x^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{e^{x^{\frac{1}{2}}}}{2 x^{\frac{1}{2}}}$

Find the Derivative – d/dx y=e^( square root of x)

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