Find the Derivative - d/da square root of ab

$\sqrt{a b}$

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

$\frac{d}{d a} [{(a b)}^{\frac{1}{2}}]$

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

To apply the Chain Rule, set $u$ as $a b$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d a} [a b]$

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d a} [a b]$

Replace all occurrences of $u$ with $a b$ .

$\frac{1}{2} {(a b)}^{\frac{1}{2} - 1} \frac{d}{d a} [a b]$

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

$\frac{1}{2} {(a b)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d a} [a b]$

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

$\frac{1}{2} {(a b)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d a} [a b]$

Combine the numerators over the common denominator.

$\frac{1}{2} {(a b)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d a} [a b]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(a b)}^{\frac{1 - 2}{2}} \frac{d}{d a} [a b]$

Subtract $2$ from $1$ .

$\frac{1}{2} {(a b)}^{\frac{- 1}{2}} \frac{d}{d a} [a b]$

Combine fractions.

Move the negative in front of the fraction.

$\frac{1}{2} {(a b)}^{- \frac{1}{2}} \frac{d}{d a} [a b]$

Combine $\frac{1}{2}$ and ${(a b)}^{- \frac{1}{2}}$ .

$\frac{{(a b)}^{- \frac{1}{2}}}{2} \frac{d}{d a} [a b]$

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

$\frac{1}{2 {(a b)}^{\frac{1}{2}}} \frac{d}{d a} [a b]$

Since $b$ is constant with respect to $a$ , the derivative of $a b$ with respect to $a$ is $b \frac{d}{d a} [a]$ .

$\frac{1}{2 {(a b)}^{\frac{1}{2}}} (b \frac{d}{d a} [a])$

Combine $b$ and $\frac{1}{2 {(a b)}^{\frac{1}{2}}}$ .

$\frac{b}{2 {(a b)}^{\frac{1}{2}}} \frac{d}{d a} [a]$

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

$\frac{b}{2 {(a b)}^{\frac{1}{2}}} \cdot 1$

Multiply $\frac{b}{2 {(a b)}^{\frac{1}{2}}}$ by $1$ .

$\frac{b}{2 {(a b)}^{\frac{1}{2}}}$

Simplify.

Apply the product rule to $a b$ .

$\frac{b}{2 (a^{\frac{1}{2}} b^{\frac{1}{2}})}$

Combine terms.

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

$\frac{b \cdot b^{- \frac{1}{2}}}{2 a^{\frac{1}{2}}}$

Multiply $b$ by $b^{- \frac{1}{2}}$ by adding the exponents.

Multiply $b$ by $b^{- \frac{1}{2}}$ .

Raise $b$ to the power of $1$ .

$\frac{b^{1} b^{- \frac{1}{2}}}{2 a^{\frac{1}{2}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{b^{1 - \frac{1}{2}}}{2 a^{\frac{1}{2}}}$

Write $1$ as a fraction with a common denominator.

$\frac{b^{\frac{2}{2} - \frac{1}{2}}}{2 a^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$\frac{b^{\frac{2 - 1}{2}}}{2 a^{\frac{1}{2}}}$

Subtract $1$ from $2$ .

$\frac{b^{\frac{1}{2}}}{2 a^{\frac{1}{2}}}$

Find the Derivative - d/da square root of ab

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