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Use the Chain Rule to find $\partial z/\partial s$ and $\partial z/\partial t$.

$z =\tan\dfrac{u}{v}$, $u = 6s + 7t$, $v = 7s − 6t$

my asnwer is

$$\frac{\partial z}{\partial s}= \frac{6\sec \left(\frac{u}{v}\right)^2}{v}+\frac{7u\sec \left(\frac{u}{v}\right)^2}{v^2}$$

$$\frac{\partial z}{\partial t}= \frac{7\sec \left(\frac{u}{v}\right)^2}{v}+\frac{6u\sec \left(\frac{u}{v}\right)^2}{v^2}$$

But the system said I am wrong. Anyone can help me check? thanks!

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    In first equation isn't derivative $\color{red}{-}\dfrac{u}{v}$ so omitted minus.? – 2017-01-28
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    It is probably easier if you set $u/v = (6s+7t)/(7s-6t) = w$ and apply the chain rule _and_ product rules (if you want the derivatives in terms of $s$ and $t$, not $u$ and $v$) \begin{align} z_{s} &= z_{w} \cdot w_{s} \\ &= \sec^{2}(w) \cdot \frac{-85t}{(7s-6t)^{2}} \\ &= \sec^{2} \left( \frac{6s+7t}{7s-6t} \right) \cdot \frac{-85t}{(7s-6t)^{2}} \end{align} – 2017-01-28
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    in addition to sign issue pointed out by MyGalsses, note $$\sec \left(\frac{u}{v}\right)^2$$ is problematic. It should be $$\sec ^2 \left(\frac{u}{v}\right)$$. – 2017-01-28
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    I am using webAssign, an online homework system, it doesn't accept your notation, that's why I used the square that way. but thanks. – 2017-01-28

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