Find the first partial derivatives of the function: $u = 9xy \sin^{-1}(yz)$

Neetesh Kumar | December 3, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 14.3 Question Number 18
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Step-by-step solution:

We are given the function: $u = 9xy \sin^{-1} (yz)$

Step 1: Find $\frac{\partial u}{\partial x}$

To compute $\frac{\partial u}{\partial x}$, treat $y$ and $z$ as constants. The derivative of $u$ with respect to $x$ is:

$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left( 9xy \sin^{-1}(yz) \right).$

Since $y \sin^{-1}(yz)$ is treated as a constant with respect to $x$, we have:

$\frac{\partial u}{\partial x} = 9y \sin^{-1}(yz).$

Step 2: Find $\frac{\partial u}{\partial y}$

To compute $\frac{\partial u}{\partial y}$, treat $x$ and $z$ as constants. Apply the product rule to:

$u = 9xy \sin^{-1}(yz).$

The product rule gives:

$\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left( 9xy \right) \sin^{-1}(yz) + 9xy \frac{\partial}{\partial y} \left( \sin^{-1}(yz) \right).$

1. The derivative of $9xy$ with respect to $y$ is $9x$.
2. The derivative of $\sin^{-1}(yz)$ with respect to $y$ is:

$\frac{\partial}{\partial y} \left( \sin^{-1}(yz) \right) = \frac{z}{\sqrt{1 - (yz)^2}}.$

Substitute back into the equation:

$\frac{\partial u}{\partial y} = 9x \sin^{-1}(yz) + 9xy \cdot \frac{z}{\sqrt{1 - (yz)^2}}.$

Simplify:

$\frac{\partial u}{\partial y} = 9x \sin^{-1}(yz) + \frac{9xyz}{\sqrt{1 - (yz)^2}}.$

Step 3: Find $\frac{\partial u}{\partial z}$

To compute $\frac{\partial u}{\partial z}$, treat $x$ and $y$ as constants. Apply the chain rule to:

$u = 9xy \sin^{-1}(yz).$

The derivative of $\sin^{-1}(yz)$ with respect to $z$ is:

$\frac{\partial}{\partial z} \left( \sin^{-1}(yz) \right) = \frac{y}{\sqrt{1 - (yz)^2}}.$

Thus:

$\frac{\partial u}{\partial z} = 9xy \cdot \frac{y}{\sqrt{1 - (yz)^2}}.$

Simplify:

$\frac{\partial u}{\partial z} = \frac{9xy^2}{\sqrt{1 - (yz)^2}}.$

Final Answer:

$\frac{\partial u}{\partial x} = \boxed{9y \sin^{-1}(yz)}$

$\frac{\partial u}{\partial y} = \boxed{9x \sin^{-1}(yz) + \frac{9xyz}{\sqrt{1 - (yz)^2}}}$

$\frac{\partial u}{\partial z} = \boxed{\frac{9xy^2}{\sqrt{1 - (yz)^2}}}$

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