PYQs / GATE CE / 2019 / Set 1 / Q40 GATE CE 2019 Set 1 — Question 40 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Partial & Total Derivatives Calculus Engineering Mathematics
Engineering Mathematics → Calculus → Partial & Total Derivatives
Last updated 5 September 2026
Question Consider two functions:
x = ψ ln ϕ x = \psi \ln \phi x = ψ ln ϕ and
y = ϕ ln ψ y = \phi \ln \psi y = ϕ ln ψ . Which one of the following is the correct expression for
∂ ψ ∂ x \frac{\partial \psi}{\partial x} ∂ x ∂ ψ ?
Correct answer (D) (ln)/(ln φ ln - 1)
Solution Differentiate both equations with respect to
x x x (treating
y y y as constant for the partial derivative
∂ ψ ∂ x \frac{\partial \psi}{\partial x} ∂ x ∂ ψ ):
1)
∂ x ∂ x = 1 = ∂ ψ ∂ x ln ϕ + ψ 1 ϕ ∂ ϕ ∂ x \frac{\partial x}{\partial x} = 1 = \frac{\partial \psi}{\partial x} \ln \phi + \psi \frac{1}{\phi} \frac{\partial \phi}{\partial x} ∂ x ∂ x = 1 = ∂ x ∂ ψ ln ϕ + ψ ϕ 1 ∂ x ∂ ϕ 2)
∂ y ∂ x = 0 = ∂ ϕ ∂ x ln ψ + ϕ 1 ψ ∂ ψ ∂ x \frac{\partial y}{\partial x} = 0 = \frac{\partial \phi}{\partial x} \ln \psi + \phi \frac{1}{\psi} \frac{\partial \psi}{\partial x} ∂ x ∂ y = 0 = ∂ x ∂ ϕ ln ψ + ϕ ψ 1 ∂ x ∂ ψ From (2), we find
∂ ϕ ∂ x = − ϕ ψ ln ψ ∂ ψ ∂ x \frac{\partial \phi}{\partial x} = -\frac{\phi}{\psi \ln \psi} \frac{\partial \psi}{\partial x} ∂ x ∂ ϕ = − ψ l n ψ ϕ ∂ x ∂ ψ .
Substitute this into (1):
1 = ∂ ψ ∂ x ln ϕ + ψ ϕ ( − ϕ ψ ln ψ ∂ ψ ∂ x ) 1 = \frac{\partial \psi}{\partial x} \ln \phi + \frac{\psi}{\phi} \left( -\frac{\phi}{\psi \ln \psi} \frac{\partial \psi}{\partial x} \right) 1 = ∂ x ∂ ψ ln ϕ + ϕ ψ ( − ψ l n ψ ϕ ∂ x ∂ ψ ) 1 = ∂ ψ ∂ x ln ϕ − 1 ln ψ ∂ ψ ∂ x 1 = \frac{\partial \psi}{\partial x} \ln \phi - \frac{1}{\ln \psi} \frac{\partial \psi}{\partial x} 1 = ∂ x ∂ ψ ln ϕ − l n ψ 1 ∂ x ∂ ψ 1 = ∂ ψ ∂ x ( ln ϕ ln ψ − 1 ln ψ ) 1 = \frac{\partial \psi}{\partial x} \left( \frac{\ln \phi \ln \psi - 1}{\ln \psi} \right) 1 = ∂ x ∂ ψ ( l n ψ l n ϕ l n ψ − 1 ) ∂ ψ ∂ x = ln ψ ln ϕ ln ψ − 1 \frac{\partial \psi}{\partial x} = \frac{\ln \psi}{\ln \phi \ln \psi - 1} ∂ x ∂ ψ = l n ϕ l n ψ − 1 l n ψ .
This matches option (D).
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Correct answer (D) (ln)/(ln φ ln - 1)
Solution Differentiate both equations with respect to
x x x (treating
y y y as constant for the partial derivative
∂ ψ ∂ x \frac{\partial \psi}{\partial x} ∂ x ∂ ψ ):
1)
∂ x ∂ x = 1 = ∂ ψ ∂ x ln ϕ + ψ 1 ϕ ∂ ϕ ∂ x \frac{\partial x}{\partial x} = 1 = \frac{\partial \psi}{\partial x} \ln \phi + \psi \frac{1}{\phi} \frac{\partial \phi}{\partial x} ∂ x ∂ x = 1 = ∂ x ∂ ψ ln ϕ + ψ ϕ 1 ∂ x ∂ ϕ 2)
∂ y ∂ x = 0 = ∂ ϕ ∂ x ln ψ + ϕ 1 ψ ∂ ψ ∂ x \frac{\partial y}{\partial x} = 0 = \frac{\partial \phi}{\partial x} \ln \psi + \phi \frac{1}{\psi} \frac{\partial \psi}{\partial x} ∂ x ∂ y = 0 = ∂ x ∂ ϕ ln ψ + ϕ ψ 1 ∂ x ∂ ψ From (2), we find
∂ ϕ ∂ x = − ϕ ψ ln ψ ∂ ψ ∂ x \frac{\partial \phi}{\partial x} = -\frac{\phi}{\psi \ln \psi} \frac{\partial \psi}{\partial x} ∂ x ∂ ϕ = − ψ l n ψ ϕ ∂ x ∂ ψ .
Substitute this into (1):
1 = ∂ ψ ∂ x ln ϕ + ψ ϕ ( − ϕ ψ ln ψ ∂ ψ ∂ x ) 1 = \frac{\partial \psi}{\partial x} \ln \phi + \frac{\psi}{\phi} \left( -\frac{\phi}{\psi \ln \psi} \frac{\partial \psi}{\partial x} \right) 1 = ∂ x ∂ ψ ln ϕ + ϕ ψ ( − ψ l n ψ ϕ ∂ x ∂ ψ ) 1 = ∂ ψ ∂ x ln ϕ − 1 ln ψ ∂ ψ ∂ x 1 = \frac{\partial \psi}{\partial x} \ln \phi - \frac{1}{\ln \psi} \frac{\partial \psi}{\partial x} 1 = ∂ x ∂ ψ ln ϕ − l n ψ 1 ∂ x ∂ ψ 1 = ∂ ψ ∂ x ( ln ϕ ln ψ − 1 ln ψ ) 1 = \frac{\partial \psi}{\partial x} \left( \frac{\ln \phi \ln \psi - 1}{\ln \psi} \right) 1 = ∂ x ∂ ψ ( l n ψ l n ϕ l n ψ − 1 ) ∂ ψ ∂ x = ln ψ ln ϕ ln ψ − 1 \frac{\partial \psi}{\partial x} = \frac{\ln \psi}{\ln \phi \ln \psi - 1} ∂ x ∂ ψ = l n ϕ l n ψ − 1 l n ψ .
This matches option (D).
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