Find dz/du z=(u^3-v^3)e^(u^3+v^3)

$z = (u^{3} - v^{3}) e^{(u^{3} + v^{3})}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d u} (z) = \frac{d}{d u} ((u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}})$

Step 2

The derivative of $z$ with respect to $u$ is $z'$ .

$z'$

Step 3

Differentiate the right side of the equation.

Differentiate using the Product Rule which states that $\frac{d}{d u} [f (u) g (u)]$ is $f (u) \frac{d}{d u} [g (u)] + g (u) \frac{d}{d u} [f (u)]$ where $f (u) = u^{3} - {z'}^{3}$ and $g (u) = e^{u^{3} + {z'}^{3}}$ .

$(u^{3} - {z'}^{3}) \frac{d}{d u} [e^{u^{3} + {z'}^{3}}] + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Differentiate using the chain rule, which states that $\frac{d}{d u} [f (g (u))]$ is $f' (g (u)) g' (u)$ where $f (u) = e^{u}$ and $g (u) = u^{3} + {z'}^{3}$ .

To apply the Chain Rule, set $u_{1}$ as $u^{3} + {z'}^{3}$ .

$(u^{3} - {z'}^{3}) (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d u} [u^{3} + {z'}^{3}]) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$(u^{3} - {z'}^{3}) (e^{u_{1}} \frac{d}{d u} [u^{3} + {z'}^{3}]) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Replace all occurrences of $u_{1}$ with $u^{3} + {z'}^{3}$ .

$(u^{3} - {z'}^{3}) (e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} + {z'}^{3}]) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Differentiate.

By the Sum Rule, the derivative of $u^{3} + {z'}^{3}$ with respect to $u$ is $\frac{d}{d u} [u^{3}] + \frac{d}{d u} [{z'}^{3}]$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (\frac{d}{d u} [u^{3}] + \frac{d}{d u} [{z'}^{3}]) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2} + \frac{d}{d u} [{z'}^{3}]) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Since ${z'}^{3}$ is constant with respect to $u$ , the derivative of ${z'}^{3}$ with respect to $u$ is $0$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2} + 0) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

Add $3 u^{2}$ and $0$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2}) + e^{u^{3} + {z'}^{3}} \frac{d}{d u} [u^{3} - {z'}^{3}]$

By the Sum Rule, the derivative of $u^{3} - {z'}^{3}$ with respect to $u$ is $\frac{d}{d u} [u^{3}] + \frac{d}{d u} [- {z'}^{3}]$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2}) + e^{u^{3} + {z'}^{3}} (\frac{d}{d u} [u^{3}] + \frac{d}{d u} [- {z'}^{3}])$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2}) + e^{u^{3} + {z'}^{3}} (3 u^{2} + \frac{d}{d u} [- {z'}^{3}])$

Since $- {z'}^{3}$ is constant with respect to $u$ , the derivative of $- {z'}^{3}$ with respect to $u$ is $0$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2}) + e^{u^{3} + {z'}^{3}} (3 u^{2} + 0)$

Simplify the expression.

Add $3 u^{2}$ and $0$ .

$(u^{3} - {z'}^{3}) e^{u^{3} + {z'}^{3}} (3 u^{2}) + e^{u^{3} + {z'}^{3}} (3 u^{2})$

Reorder terms.

$3 e^{u^{3} + {z'}^{3}} u^{2} (u^{3} - {z'}^{3}) + 3 e^{u^{3} + {z'}^{3}} u^{2}$

Step 4

Reform the equation by setting the left side equal to the right side.

$z' = 3 e^{u^{3} + {z'}^{3}} u^{2} (u^{3} - {z'}^{3}) + 3 e^{u^{3} + {z'}^{3}} u^{2}$

Step 5

Simplify.

Simplify each term.

Apply the distributive property.

$z' = 3 e^{u^{3} + {z'}^{3}} u^{2} u^{3} + 3 e^{u^{3} + {z'}^{3}} u^{2} (- {z'}^{3}) + 3 e^{u^{3} + {z'}^{3}} u^{2}$

Multiply $u^{2}$ by $u^{3}$ by adding the exponents.

Move $u^{3}$ .

$z' = 3 e^{u^{3} + {z'}^{3}} (u^{3} u^{2}) + 3 e^{u^{3} + {z'}^{3}} u^{2} (- {z'}^{3}) + 3 e^{u^{3} + {z'}^{3}} u^{2}$

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

$z' = 3 e^{u^{3} + {z'}^{3}} u^{3 + 2} + 3 e^{u^{3} + {z'}^{3}} u^{2} (- {z'}^{3}) + 3 e^{u^{3} + {z'}^{3}} u^{2}$

Add $3$ and $2$ .

$z' = 3 e^{u^{3} + {z'}^{3}} u^{5} + 3 e^{u^{3} + {z'}^{3}} u^{2} (- {z'}^{3}) + 3 e^{u^{3} + {z'}^{3}} u^{2}$

Multiply $- 1$ by $3$ .

$z' = 3 e^{u^{3} + {z'}^{3}} u^{5} - 3 e^{u^{3} + {z'}^{3}} u^{2} {z'}^{3} + 3 e^{u^{3} + {z'}^{3}} u^{2}$

Reorder factors in $3 e^{u^{3} + {z'}^{3}} u^{5} - 3 e^{u^{3} + {z'}^{3}} u^{2} {z'}^{3} + 3 e^{u^{3} + {z'}^{3}} u^{2}$ .

$z' = 3 u^{5} e^{u^{3} + {z'}^{3}} - 3 u^{2} e^{u^{3} + {z'}^{3}} {z'}^{3} + 3 u^{2} e^{u^{3} + {z'}^{3}}$

Step 6

Replace $z'$ with $\frac{d z}{d u}$ .

$\frac{d z}{d u} = 3 u^{5} e^{u^{3} + {z'}^{3}} - 3 u^{2} e^{u^{3} + {z'}^{3}} {z'}^{3} + 3 u^{2} e^{u^{3} + {z'}^{3}}$

Find dz/du z=(u^3-v^3)e^(u^3+v^3)

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