# Thread: Derivation of n functions

1. ## Derivation of n functions

Calculate

1) $\displaystyle z=e^\frac{s}{t}$, $\displaystyle s=xy^2$ and $\displaystyle t=5x+2y^3$. Calculate $\displaystyle \frac{dz}{dx}$ and $\displaystyle \frac{dz}{dy}$

My solution:

$\displaystyle \frac{dz}{dx} = \frac{dz}{ds} \frac{ds}{dx} + \frac{dz}{dt} \frac{dt}{dx}$

$\displaystyle \frac{dz}{dx}= \frac{y^2e^\frac{s}{t}}{t}-\frac{5se^\frac{s}{t}}{t^2}$

$\displaystyle \frac{dz}{dy} = \frac{2xye^\frac{s}{t}}{t}-\frac{6yse^\frac{s}{t}}{t^2}$????

2. Originally Posted by Apprentice123
Calculate

1) $\displaystyle z=e^\frac{s}{t}$, $\displaystyle s=xy^2$ and $\displaystyle t=5x+2y^3$. Calculate $\displaystyle \frac{dz}{dx}$ and $\displaystyle \frac{dz}{dy}$

My solution:

$\displaystyle \frac{dz}{dx} = \frac{dz}{ds} \frac{ds}{dx} + \frac{dz}{dt} \frac{dt}{dx}$

$\displaystyle \frac{dz}{dx}= \frac{y^2e^\frac{s}{t}}{t}-\frac{5se^\frac{s}{t}}{t^2}$

$\displaystyle \frac{dz}{dy} = \frac{2xye^\frac{s}{t}}{t}-\frac{6yse^\frac{s}{t}}{t^2}$???? y should be y^2 in the second fraction. Otherwise it's correct.