Find the Derivative – d/dy e^(x/y)

$e^{\frac{x}{y}}$

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

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

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

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

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

Replace all occurrences of $u$ with $\frac{x}{y}$ .

$e^{\frac{x}{y}} \frac{d}{d y} [\frac{x}{y}]$

Differentiate.

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Since $x$ is constant with respect to $y$ , the derivative of $\frac{x}{y}$ with respect to $y$ is $x \frac{d}{d y} [\frac{1}{y}]$ .

$e^{\frac{x}{y}} (x \frac{d}{d y} [\frac{1}{y}])$

Rewrite $\frac{1}{y}$ as $y^{- 1}$ .

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

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

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

Simplify the expression.

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Move $- 1$ to the left of $e^{\frac{x}{y}} x$ .

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

Rewrite $- 1 e^{\frac{x}{y}}$ as $- e^{\frac{x}{y}}$ .

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

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Combine terms.

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

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

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

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

Find the Derivative – d/dy e^(x/y)

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