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Calculus IIIChapter 2: Partial DerivativesSection 2-6

Chain Rule — Question 3

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Question 3

Let w=x2+yz,x=s+t,y=st,z=s2−t2.w=x^2+yz,\quad x=s+t,\quad y=st,\quad z=s^2-t^2. Tasks

  1. Draw or use the dependency structure to find ws,wtw_s,w_t.

  2. Evaluate both at (s,t)=(1,2)(s,t)=(1,2).

  3. Verify one result by explicit expansion.

Original worksheet page 1: question and worked solution for 2-6-003
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Question 3 – Solution

Strategy. Sum the contribution along every dependency path.

See the diagram in the original worksheet below.

Step 1: Derivatives wx=2x,wy=z,wz=y.w_x=2x,\quad w_y=z,\quad w_z=y. xs=1,ys=t,zs=2s;xt=1,yt=s,zt=−2t.x_s=1,\ y_s=t,\ z_s=2s;\qquad x_t=1,\ y_t=s,\ z_t=-2t. Therefore ws=2x+zt+2sy,wt=2x+zs−2ty.\boxed{w_s=2x+zt+2sy},\qquad \boxed{w_t=2x+zs-2ty}.

Step 2: Evaluate At (1,2)(1,2), (x,y,z)=(3,2,−3)(x,y,z)=(3,2,-3): ws=6−6+4=4,wt=6−3−8=−5.\boxed{w_s=6-6+4=4},\qquad\boxed{w_t=6-3-8=-5}.

Step 3: Check Expanding gives w=(s+t)2+st(s2−t2)w=(s+t)^2+st(s^2-t^2). Differentiating in ss yields 2(s+t)+t(3s2−t2)2(s+t)+t(3s^2-t^2), which equals 44 at (1,2)(1,2).

Original worksheet page 2: question and worked solution for 2-6-003

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