Question

Find \(d w / d t\) using the appropriate Chain Rule. \(\begin{aligned} w &=x \sec y \\ x &=e^{t}, \quad y=\pi-t \end{aligned}\)

Short answer

The derivative \(dw/dt = \sec (\pi - t) * e^t - e^t \sec (\pi - t) \tan (\pi - t)\)

Step-by-step solution

Find dx/dt

Here, \(x = e^t\). So, \(dx/dt = e^t\)

Find dy/dt

In this case, \(y = \pi - t\). Therefore, \(dy/dt = -1\) since derivative of \(t\) is 1 and derivative of a constant \(\pi\) is 0.

Find dw/dx

The partial derivative of \(w\) with respect to \(x\) is \(\sec y\), as \(w = x \sec y\). So, \(dw/dx = \sec y\).

Find dw/dy

The partial derivative of \(w\) with respect to \(y\) involves differentiating \(\sec y\) with respect to \(y\), and multiplying by the factor \(x\). Since the derivative of \(\sec y\) is \(\sec y \tan y\), we have \(dw/dy = x \sec y \tan y\).

Apply Chain Rule

With the four derivatives found, apply the Chain Rule \(dw/dt = dw/dx * dx/dt + dw/dy * dy/dt\). Substituting the derivatives into it, \(dw/dt = \sec y * e^t + x \sec y \tan y * -1\)

Replace x and y

To find the final result, replace values of \(x\) and \(y\) back in terms of \(t\). So, \(dw/dt = \sec (\pi - t) * e^t + e^t \sec (\pi - t) \tan (\pi - t) * -1\).