Use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$: $x^9 + y^3 + z^8 = 5xyz$

Neetesh Kumar | December 3, 2024

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This is the solution to Math 1D
Assignment: 14.3 Question Number 21
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Step-by-step solution:

We are asked to differentiate the equation implicitly to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. We will differentiate both sides of the equation with respect to $x$ and $y$, treating $z$ as a function of $x$ and $y$.

Step 1: Differentiate with respect to $x$

Differentiate the equation $x^9 + y^3 + z^8 = 5xyz$ with respect to $x$, keeping in mind that $z$ is a function of $x$.

$\frac{\partial}{\partial x} \left( x^9 + y^3 + z^8 \right) = \frac{\partial}{\partial x} \left( 5xyz \right)$

• The derivative of $x^9$ with respect to $x$ is $9x^8$.
• The derivative of $y^3$ with respect to $x$ is $0$, since $y$ is independent of $x$.
• Using the chain rule, the derivative of $z^8$ with respect to $x$ is $8z^7 \frac{\partial z}{\partial x}$.
• Applying the product rule to $5xyz$, we get: $\frac{\partial}{\partial x} \left( 5xyz \right) = 5yz + 5xz \frac{\partial y}{\partial x} + 5xy \frac{\partial z}{\partial x}$

Thus, we get the equation:

$9x^8 + 8z^7 \frac{\partial z}{\partial x} = 5yz + 5xy \frac{\partial z}{\partial x}$

Now, solve for $\frac{\partial z}{\partial x}$:

$8z^7 \frac{\partial z}{\partial x} - 5xy \frac{\partial z}{\partial x} = 5yz - 9x^8$

Factor out $\frac{\partial z}{\partial x}$:

$\frac{\partial z}{\partial x} \left( 8z^7 - 5xy \right) = 5yz - 9x^8$

Finally, solve for $\frac{\partial z}{\partial x}$:

$\frac{\partial z}{\partial x} = \frac{5yz - 9x^8}{8z^7 - 5xy}$

Step 2: Differentiate with respect to $y$

Now, differentiate the original equation $x^9 + y^3 + z^8 = 5xyz$ with respect to $y$.

$\frac{\partial}{\partial y} \left( x^9 + y^3 + z^8 \right) = \frac{\partial}{\partial y} \left( 5xyz \right)$

• The derivative of $x^9$ with respect to $y$ is $0$, since $x$ is independent of $y$.
• The derivative of $y^3$ with respect to $y$ is $3y^2$.
• Using the chain rule, the derivative of $z^8$ with respect to $y$ is $8z^7 \frac{\partial z}{\partial y}$.
• Applying the product rule to $5xyz$, we get: $\frac{\partial}{\partial y} \left( 5xyz \right) = 5xz + 5xy \frac{\partial z}{\partial y}$

Thus, we get the equation:

$3y^2 + 8z^7 \frac{\partial z}{\partial y} = 5xz + 5xy \frac{\partial z}{\partial y}$

Now, solve for $\frac{\partial z}{\partial y}$:

$8z^7 \frac{\partial z}{\partial y} - 5xy \frac{\partial z}{\partial y} = 5xz - 3y^2$

Factor out $\frac{\partial z}{\partial y}$:

$\frac{\partial z}{\partial y} \left( 8z^7 - 5xy \right) = 5xz - 3y^2$

Finally, solve for $\frac{\partial z}{\partial y}$:

$\frac{\partial z}{\partial y} = \frac{5xz - 3y^2}{8z^7 - 5xy}$

Final Answers:

$\frac{\partial z}{\partial x} = \boxed{\frac{5yz - 9x^8}{8z^7 - 5xy}}$

$\frac{\partial z}{\partial y} = \boxed{\frac{5xz - 3y^2}{8z^7 - 5xy}}$

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