# Thread: Differentiation! Please help!

1. ## Differentiation! Please help!

I'm working on Survey of Calculus.
Please help!

Evaluate dy/dt for the function at the point.

x^3 e^y - y^3 ln x = 7; dx/dt = 2, x = 1, y = 3

2. Hello, Matho!

Evaluate $\frac{dy}{dt}$ for the function at the point.
. . . $x^3e^y - y^3\ln x \:=\: 7,\quad x=1,\;y=3,\;\frac{dx}{dt} = 2$
Differentiate implicitly ... (Don't forget the Product Rule!)

. . $x^3e^y\,\frac{dy}{dt} + 3x^2e^y\,\frac{dx}{dt} - y^3\,\frac{1}{x}\,\frac{dx}{dt} - 3y^2\ln x\,\frac{dy}{dt} \;=\;0$

Substitute the given values:

. . $(1^3)(e^3)\,\frac{dy}{dt} + (3)(1^2)(e^3)(2) - (3^3)\left(\tfrac{1}{1}\right)(2) - (3)(3^2)(\ln 1)\,\frac{dy}{dt} \;=\;0$

. . $e^3\!\cdot\!\frac{dy}{dt} + 6e^3 - 54 - 27\!\cdot\!0\!\cdot\!\frac{dy}{dt} \;=\;0$

. . $e^3\!\cdot\!\frac{dy}{dt} \;=\;54 - 6e^3$

. . $\frac{dy}{dt} \;=\;\frac{6(9-e^3)}{e^3}$