2070 Magh · 5 marks

Total derivative of sin(x/y)

Short answer

Using chain rule, du/dt = (1/y)cos(x/y)·e^t − (x/y²)cos(x/y)·2t. With x=e^t, y=t², du/dt = e^t (t−2) cos(e^t/t²) / t³.

Find du/dt if u = sin(x/y), x = eᵗ and y = t².

Here u = sin(x/y) is a function of x and y, which are themselves functions of t. By the chain rule for total derivatives:

du/dt = (∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt)

Partial derivatives

∂u/∂x = cos(x/y) · (1/y)      = cos(x/y)/y
∂u/∂y = cos(x/y) · (−x/y²)    = −x cos(x/y)/y²

Derivatives of x and y

x = eᵗ → dx/dt = eᵗ; y = t² → dy/dt = 2t.

Substitute

du/dt = cos(x/y)/y · eᵗ  +  ( −x cos(x/y)/y² ) · 2t
      = cos(eᵗ/t²) [ eᵗ/t²  −  eᵗ · 2t / t⁴ ]
      = cos(eᵗ/t²) [ eᵗ/t²  −  2eᵗ/t³ ]
      = eᵗ (t − 2)/t³ · cos(eᵗ/t²)

Answer: du/dt = (t − 2) eᵗ cos(eᵗ/t²) / t³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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