Use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$: $x - z = 7 \arctan(yz)$

Neetesh Kumar | December 3, 2024

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This is the solution to Math 1D
Assignment: 14.3 Question Number 22
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Step-by-step solution:

Step 1: Differentiate implicitly with respect to $x$

Take the partial derivative of both sides of the equation $x - z = 7 \arctan(yz)$ with respect to $x$:

$\frac{\partial}{\partial x}(x - z) = \frac{\partial}{\partial x}\left(7 \arctan(yz)\right).$

• The derivative of $x$ with respect to $x$ is $1$.

• The derivative of $z$ with respect to $x$ is $\frac{\partial z}{\partial x}$, so the left-hand side becomes: $1 - \frac{\partial z}{\partial x}.$

• For the right-hand side, use the chain rule for $\arctan(u)$, where $u = yz$. The derivative of $\arctan(u)$ is $\frac{1}{1 + u^2}$, and apply the product rule to $yz$:

  $\frac{\partial}{\partial x}\left(7 \arctan(yz)\right) = 7 \cdot \frac{1}{1 + (yz)^2} \cdot \frac{\partial}{\partial x}(yz).$

Since $y$ is independent of $x$, the derivative of $yz$ with respect to $x$ is:

$\frac{\partial}{\partial x}(yz) = y \cdot \frac{\partial z}{\partial x}.$

Thus: $\frac{\partial}{\partial x}\left(7 \arctan(yz)\right) = \frac{7y \cdot \frac{\partial z}{\partial x}}{1 + (yz)^2}.$

Equating the derivatives: $1 - \frac{\partial z}{\partial x} = \frac{7y \cdot \frac{\partial z}{\partial x}}{1 + (yz)^2}.$

Step 2: Solve for $\frac{\partial z}{\partial x}$

Rearrange to isolate $\frac{\partial z}{\partial x}$: $1 = \frac{\partial z}{\partial x} \left(1 + \frac{7y}{1 + (yz)^2}\right).$

Factor out $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x} = \frac{1}{1 + \frac{7y}{1 + (yz)^2}}.$

Simplify: $\frac{\partial z}{\partial x} = \frac{1 + (yz)^2}{1 + (yz)^2 + 7y}.$

Step 3: Differentiate implicitly with respect to $y$

Take the partial derivative of both sides of $x - z = 7 \arctan(yz)$ with respect to $y$:

$\frac{\partial}{\partial y}(x - z) = \frac{\partial}{\partial y}\left(7 \arctan(yz)\right).$

• The derivative of $x$ with respect to $y$ is $0$ (since $x$ is independent of $y$).

• The derivative of $-z$ with respect to $y$ is $-\frac{\partial z}{\partial y}$.

For the right-hand side, differentiate $7 \arctan(yz)$ with respect to $y$:

$\frac{\partial}{\partial y}\left(7 \arctan(yz)\right) = 7 \cdot \frac{1}{1 + (yz)^2} \cdot \frac{\partial}{\partial y}(yz).$

The derivative of $yz$ with respect to $y$ is: $\frac{\partial}{\partial y}(yz) = z + y \cdot \frac{\partial z}{\partial y}.$

Thus: $\frac{\partial}{\partial y}\left(7 \arctan(yz)\right) = \frac{7(z + y \cdot \frac{\partial z}{\partial y})}{1 + (yz)^2}.$

Equating the derivatives: $-\frac{\partial z}{\partial y} = \frac{7(z + y \cdot \frac{\partial z}{\partial y})}{1 + (yz)^2}.$

Step 4: Solve for $\frac{\partial z}{\partial y}$

Rearrange to isolate $\frac{\partial z}{\partial y}$:

$-\frac{\partial z}{\partial y} \left(1 + \frac{7y}{1 + (yz)^2}\right) = \frac{7z}{1 + (yz)^2}.$

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

$\frac{\partial z}{\partial y} = -\frac{\frac{7z}{1 + (yz)^2}}{1 + \frac{7y}{1 + (yz)^2}}.$

Simplify: $\frac{\partial z}{\partial y} = -\frac{7z}{1 + (yz)^2 + 7y}.$

Final Answers:

$\frac{\partial z}{\partial x} = \boxed{\frac{1 + (yz)^2}{1 + (yz)^2 + 7y}}$

$\frac{\partial z}{\partial y} = \boxed{-\frac{7z}{1 + (yz)^2 + 7y}}$

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