Question

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=\cos (x-y), \quad x=t^{2}, \quad y=1\)

Short answer

The derivative of the function with respect to \(t\), \(\displaystyle\frac{{dw}}{{dt}} = -2t\sin(t^{2}-1)\).

Step-by-step solution

Recognize the appropriate Chain Rule

Given that \(w=\cos (x-y)\) where \(x=t^{2}\) and \(y=1\). The chain rule states \(\displaystyle\frac{{dw}}{{dt}}=\frac{{dw}}{{dx}}\frac{{dx}}{{dt}}-\frac{{dw}}{{dy}}\frac{{dy}}{{dt}}\). Polynomial differential rules will also be applied.

Differentiate using the Chain Rule

\[\frac{{dw}}{{dx}} = -\sin(x-y)\], \[\frac{{dx}}{{dt}} = 2t \] and since \(y\) is a constant, \[\frac{{dw}}{{dy}} = 0\], \[\frac{{dy}}{{dt}} = 0.\] Substituting these expressions into the chain rule gives: \[\frac{{dw}}{{dt}} = -\sin(x-y)\cdot2t - 0\cdot0 = -2t\sin(t^{2}-1).\]

Convert \(w\) to a function of \(t\) before differentiating

Replace \(x\) and \(y\) with their respective values in terms of \(t\) in original function to yield: \(w = \cos(t^{2}-1)\).

Differentiate after conversion

Differentiate \(w\) with respect to \(t\) as follows: \[\displaystyle\frac{{dw}}{{dt}} = -\sin(t^{2}-1)\cdot2t = -2t\sin(t^{2}-1).\]

Verify the answers from both methods match

The results from steps 2 and 4 should coincide. In this case, they both do: \[\displaystyle\frac{{dw}}{{dt}} = -2t\sin(t^{2}-1).\] Hence, both methods are correct