Question

If \(f = \text{Tan}^{-1}(xy)\) then \((\frac{\partial f}{\partial x})_{(1,2)}\) = _____ .

• A. \( \frac{1}{5} \)
• B. \( \frac{2}{5} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{4}{5} \)

Hint:

Partial Derivatives. When differentiating with respect to one variable (e.g., x), treat other variables (e.g., y) as constants. Use standard differentiation rules (like chain rule). Evaluate the resulting expression at the given point. \( \frac{d{du(\arctan u) = \frac{1{1+u^2 \).

Answer 1

The Correct Option is B

We need to find the partial derivative of \( f(x, y) = \arctan(xy) \) with respect to \(x\), and then evaluate it at the point (1, 2)
Recall the derivative of arctan(u): \( \frac{d}{dx}(\arctan u) = \frac{1}{1+u^2} \frac{du}{dx} \)
Here, treat \(y\) as a constant when differentiating with respect to \(x\)
Let \(u = xy\)
Then \( \frac{\partial u}{\partial x} = y \)
$$ \frac{\partial f}{\partial x} = \frac{1}{1+(xy)^2} \cdot \frac{\partial (xy)}{\partial x} $$ $$ \frac{\partial f}{\partial x} = \frac{1}{1+x^2y^2} \cdot y = \frac{y}{1+x^2y^2} $$ Now evaluate this partial derivative at the point (x=1, y=2): $$ \left(\frac{\partial f}{\partial x}\right)_{(1,2)} = \frac{2}{1+(1)^2(2)^2} = \frac{2}{1 + 1 \times 4} = \frac{2}{1+4} = \frac{2}{5} $$