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I've been given the following:

$$ \begin{cases} z = \ln(x + y^2)\\ x = t + 1/s\\ y = t \end{cases} $$

and have been asked to find $\frac{\partial z}{\partial t}$, $\frac{\partial z}{\partial s}$, and $\frac{\partial^2z}{\partial s \partial t}$.

I've calculated that:

$$ \frac{\partial z}{\partial t} = \frac{1 + 2t}{t + 1/s + t^2} $$

and that

$$ \frac{\partial z}{\partial s} = -\frac{1}{s^2(t + 1/s + t^2)} $$

This is fine and matches the mark scheme I've been given. However I cannot seem to work out $\frac{\partial^2z}{\partial s \partial t}$. I was under the impression that:

$$ \frac{\partial^2z}{\partial s \partial t} = \frac{\partial z}{\partial t}\frac{\partial z}{\partial s} $$

Is that wrong? And if so what is the correct way of calculating this?

Fabrosi
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George Baker
  • 13

1 Answers1

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The expression $\frac{\partial^2 z}{\partial s\partial t}$ means: $$\frac{\partial^2z}{\partial s\partial t}=\frac{\partial }{\partial s}\left(\frac{\partial z}{\partial t}\right)$$ Then, \begin{align*} \frac{\partial^2 z}{\partial s\partial t}&=\frac{\partial}{\partial s}\left(\frac{1+2t}{t+1/s+t^2}\right)\\ &=\frac{\left[\frac{\partial}{\partial s}(1+2t)\right](t+1/s+t^2)-(1+2t)\left[\frac{\partial}{\partial s}(t+1/s+t^2)\right]}{(t+1/s+t^2)^2}\\ &=\frac{(0)(t+1/s+t^2)-(1+2t)\left(-\frac{1}{s^2}\right)}{(t+1/s+t^2)^2}\\ &=\frac{1+2t}{\left[1+s(t+t^2)\right]^2} \end{align*}

Ángel Mario Gallegos
  • 19,882