GATE 2020 EC Q.3: solution and answer

GATE 2020 · ECQ.3 · 1 mark · MCQ

The partial derivative of the function

f (x, y, z) = e^{1 - x c o s y} + x z e^{- 1/ (1 + y^{2})}

with respect to $x$ at the point $(1, 0, e)$ is

(A)
$- 1$
(B)
$0$
(C)
$1$
(D)
$\frac{1}{e}$

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Answer and solution

ANSWER

$\frac{\partial f}{\partial x}_{(1, 0, e)} = - 1 + 1 = 0$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2020 EC (IIT Delhi)

Differentiate partially with respect to $x$ , treating $y$ and $z$ as constants.
First term, by the chain rule: $\frac{\partial}{\partial x} e^{1 - x c o s y} = e^{1 - x c o s y} \cdot (- cos y) = - cos y e^{1 - x c o s y}$ .
Second term: the factor $z e^{- 1/ (1 + y^{2})}$ does not depend on $x$ , so $\frac{\partial}{\partial x} (x z e^{- 1/ (1 + y^{2})}) = z e^{- 1/ (1 + y^{2})}$ .
Adding the two terms: $\frac{\partial f}{\partial x} = - cos y e^{1 - x c o s y} + z e^{- 1/ (1 + y^{2})} .$
Substitute $(x, y, z) = (1, 0, e)$ . Since $cos 0 = 1$ , $e^{1 - 1 \cdot 1} = e^{0} = 1$ . Also $e^{- 1/ (1 + 0^{2})} = e^{- 1}$ .
$\frac{\partial f}{\partial x}_{(1, 0, e)} = - 1 \cdot 1 + e \cdot e^{- 1} = - 1 + 1 = 0.$
The value is $0$ , which is option (B).

KEY IDEA

In a partial derivative with respect to $x$ , treat $y$ and $z$ as constants. The chain rule gives the exponential term the factor $- cos y$ , and the second term is linear in $x$ , so its derivative is just its coefficient $z e^{- 1/ (1 + y^{2})}$ . Differentiate in general first, then substitute the point.

WATCH OUT FOR

Leaving out the chain-rule factor $- cos y$ on the first term. That gives $e^{0} + e \cdot e^{- 1} = 2$ , which is not an option, or, with the sign slipped, a wrong answer such as $1$ . The other common slip is substituting $x = 1$ before differentiating, which removes the $x$ -dependence you are meant to differentiate.

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GATE 2020 EC Q.3: solution and answer

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