Partial Derivatives — Question 5

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Question 5

Let F(x,y,z)=x2yz+ln⁡(x+z)−yez,x+z>0.F(x,y,z)=x^2yz+\ln(x+z)-ye^z,\qquad x+z>0. Tasks

  1. Compute FxF_x, FyF_y, and FzF_z.

  2. Evaluate all three at (1,2,0)(1,2,0).

  3. Check which terms are constant in each differentiation.

Original worksheet page 1: question and worked solution for 2-2-005
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Question 5 – Solution

Strategy. Differentiate with one variable active at a time and keep the logarithm’s domain restriction.

Step 1: Partial derivatives Fx=2xyz+1x+z,\boxed{F_x=2xyz+\frac 1{x+z}}, Fy=x2z−ez,\boxed{F_y=x^2z-e^z}, Fz=x2y+1x+z−yez.\boxed{F_z=x^2y+\frac 1{x+z}-ye^z}.

Step 2: Evaluate At (1,2,0)(1,2,0), Fx=0+1=1,Fy=0−1=−1,F_x=0+1=\boxed 1,\qquad F_y=0-1=\boxed{-1}, Fz=2+1−2=1.F_z=2+1-2=\boxed 1.

Verification. In FxF_x, −yez-ye^z is constant; in FyF_y, ln⁡(x+z)\ln(x+z) is constant; in FzF_z, every term varies with zz. The point satisfies x+z=1>0x+z=1>0.

Original worksheet page 2: question and worked solution for 2-2-005

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