Question

Find df/dt at t = 0, where
f (x, y) = x cos y + e^x sin y, x = t^2 + 1, y = t^3 + t.​

Answer 1

Answer:

$\frac{df}{dx} = 1$

Step-by-step explanation:

$\frac{df}{dx} = \frac{d(x \cos(y) + {e}^{x} )}{dx} \\ \\ = (\cos(y) - x \sin(y) \frac{dy}{dx} ) \\ \\ \frac{dy}{dt} = 1 + 3 {t}^{2} \\ \\ \frac{dx}{dx} =2 t \\ \\ \frac{dy}{dx} = \frac{1 + 3 {t}^{2} }{2t} \\ \\ \frac{dy}{dx} = (lim \ \: \: \: t\: \: \: \: tends \: to \: 0) \: \: \: \: \: \: \frac{1 + 3 {t}^{2} }{2t} = \frac{6t}{2} (by \: l \: hospital \: rule) \\ \\ \frac{dy}{dx} = 0 \\ \\ \\ \frac{df}{dx} = (\cos(y) - x \sin(y) \frac{dy}{dx} ) \\ \\ at \: \: t \: = 0 \: \: y = 0 \\ \\ at \: t = 0 \: \: x = 1 \\ \\ \\ \\ \frac{df}{dx} = ( \cos(0) - 1 \sin(0) (0)) \\ \\ \frac{df}{dx} = 1$