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**Belowzero78** Let $\displaystyle z =f(x,y) = (8+25x)y^{tan(x)}) $ , where: $\displaystyle x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t) $, and $\displaystyle y = (2 - t)e^{t} - 12(t-1) $. Use the chain rule to find $\displaystyle \frac{dz}{dt} $ at $\displaystyle t = 1 $.

$\displaystyle \frac{dz}{dt} = ?$

The Chain rule formula I used is the following: $\displaystyle \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} $

In your first partial, you should an x wrt t not y wrt t but you did the correct derivatives in your work.

And, when $\displaystyle t = 1 $, $\displaystyle x = \frac{\pi}{4} $

When $\displaystyle t = 1 $, $\displaystyle y = e $

$\displaystyle \frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)} $

$\displaystyle \frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1} $. This is one correct?

$\displaystyle \frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t} $

$\displaystyle \frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}} $