PYQs / GATE EC / 2020 / Set 1 / Q13 GATE EC 2020 Set 1 — Question 13 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Easy Partial & Total Derivatives Calculus Engineering Mathematics
Engineering Mathematics → Calculus → Partial & Total Derivatives
Last updated 5 September 2026
Question The partial derivative of the function
f ( x , y , z ) = e 1 − x cos y + x z e − 1 / ( 1 + y 2 ) f(x, y, z) = e^{1-x \cos y} + xze^{-1/(1+y^2)} f ( x , y , z ) = e 1 − x c o s y + x z e − 1/ ( 1 + y 2 ) with respect to
x x x at the point
( 1 , 0 , e ) (1, 0, e) ( 1 , 0 , e ) is
Solution Given the function
f ( x , y , z ) = e 1 − x cos y + x z e − 1 / ( 1 + y 2 ) f(x, y, z) = e^{1-x \cos y} + xze^{-1/(1+y^2)} f ( x , y , z ) = e 1 − x c o s y + x z e − 1/ ( 1 + y 2 ) .
Step 1: Find the partial derivative with respect to
x x x :
∂ f ∂ x = ∂ ∂ x ( e 1 − x cos y ) + ∂ ∂ x ( x z e − 1 / ( 1 + y 2 ) ) \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(e^{1-x \cos y}\right) + \frac{\partial}{\partial x}\left(xze^{-1/(1+y^2)}\right) ∂ x ∂ f = ∂ x ∂ ( e 1 − x c o s y ) + ∂ x ∂ ( x z e − 1/ ( 1 + y 2 ) ) ∂ f ∂ x = e 1 − x cos y ⋅ ( − cos y ) + z e − 1 / ( 1 + y 2 ) \frac{\partial f}{\partial x} = e^{1-x \cos y} \cdot (-\cos y) + ze^{-1/(1+y^2)} ∂ x ∂ f = e 1 − x c o s y ⋅ ( − cos y ) + z e − 1/ ( 1 + y 2 ) Step 2: Evaluate at the point
( 1 , 0 , e ) (1, 0, e) ( 1 , 0 , e ) , where
x = 1 , y = 0 , z = e x=1, y=0, z=e x = 1 , y = 0 , z = e :
∂ f ∂ x ∣ ( 1 , 0 , e ) = e 1 − 1 ⋅ cos ( 0 ) ⋅ ( − cos ( 0 ) ) + e ⋅ e − 1 / ( 1 + 0 2 ) \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = e^{1-1 \cdot \cos(0)} \cdot (-\cos(0)) + e \cdot e^{-1/(1+0^2)} ∂ x ∂ f ( 1 , 0 , e ) = e 1 − 1 ⋅ c o s ( 0 ) ⋅ ( − cos ( 0 )) + e ⋅ e − 1/ ( 1 + 0 2 ) Since
cos ( 0 ) = 1 \cos(0) = 1 cos ( 0 ) = 1 :
∂ f ∂ x ∣ ( 1 , 0 , e ) = e 1 − 1 ⋅ ( − 1 ) + e ⋅ e − 1 \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = e^{1-1} \cdot (-1) + e \cdot e^{-1} ∂ x ∂ f ( 1 , 0 , e ) = e 1 − 1 ⋅ ( − 1 ) + e ⋅ e − 1 ∂ f ∂ x ∣ ( 1 , 0 , e ) = e 0 ⋅ ( − 1 ) + e 1 ⋅ e − 1 \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = e^0 \cdot (-1) + e^1 \cdot e^{-1} ∂ x ∂ f ( 1 , 0 , e ) = e 0 ⋅ ( − 1 ) + e 1 ⋅ e − 1 ∂ f ∂ x ∣ ( 1 , 0 , e ) = − 1 + 1 = 0 \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = -1 + 1 = 0 ∂ x ∂ f ( 1 , 0 , e ) = − 1 + 1 = 0 Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Solution Given the function
f ( x , y , z ) = e 1 − x cos y + x z e − 1 / ( 1 + y 2 ) f(x, y, z) = e^{1-x \cos y} + xze^{-1/(1+y^2)} f ( x , y , z ) = e 1 − x c o s y + x z e − 1/ ( 1 + y 2 ) .
Step 1: Find the partial derivative with respect to
x x x :
∂ f ∂ x = ∂ ∂ x ( e 1 − x cos y ) + ∂ ∂ x ( x z e − 1 / ( 1 + y 2 ) ) \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(e^{1-x \cos y}\right) + \frac{\partial}{\partial x}\left(xze^{-1/(1+y^2)}\right) ∂ x ∂ f = ∂ x ∂ ( e 1 − x c o s y ) + ∂ x ∂ ( x z e − 1/ ( 1 + y 2 ) ) ∂ f ∂ x = e 1 − x cos y ⋅ ( − cos y ) + z e − 1 / ( 1 + y 2 ) \frac{\partial f}{\partial x} = e^{1-x \cos y} \cdot (-\cos y) + ze^{-1/(1+y^2)} ∂ x ∂ f = e 1 − x c o s y ⋅ ( − cos y ) + z e − 1/ ( 1 + y 2 ) Step 2: Evaluate at the point
( 1 , 0 , e ) (1, 0, e) ( 1 , 0 , e ) , where
x = 1 , y = 0 , z = e x=1, y=0, z=e x = 1 , y = 0 , z = e :
∂ f ∂ x ∣ ( 1 , 0 , e ) = e 1 − 1 ⋅ cos ( 0 ) ⋅ ( − cos ( 0 ) ) + e ⋅ e − 1 / ( 1 + 0 2 ) \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = e^{1-1 \cdot \cos(0)} \cdot (-\cos(0)) + e \cdot e^{-1/(1+0^2)} ∂ x ∂ f ( 1 , 0 , e ) = e 1 − 1 ⋅ c o s ( 0 ) ⋅ ( − cos ( 0 )) + e ⋅ e − 1/ ( 1 + 0 2 ) Since
cos ( 0 ) = 1 \cos(0) = 1 cos ( 0 ) = 1 :
∂ f ∂ x ∣ ( 1 , 0 , e ) = e 1 − 1 ⋅ ( − 1 ) + e ⋅ e − 1 \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = e^{1-1} \cdot (-1) + e \cdot e^{-1} ∂ x ∂ f ( 1 , 0 , e ) = e 1 − 1 ⋅ ( − 1 ) + e ⋅ e − 1 ∂ f ∂ x ∣ ( 1 , 0 , e ) = e 0 ⋅ ( − 1 ) + e 1 ⋅ e − 1 \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = e^0 \cdot (-1) + e^1 \cdot e^{-1} ∂ x ∂ f ( 1 , 0 , e ) = e 0 ⋅ ( − 1 ) + e 1 ⋅ e − 1 ∂ f ∂ x ∣ ( 1 , 0 , e ) = − 1 + 1 = 0 \left. \frac{\partial f}{\partial x} \right|_{(1, 0, e)} = -1 + 1 = 0 ∂ x ∂ f ( 1 , 0 , e ) = − 1 + 1 = 0 Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
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