Partial derivative

**mr_motivator** · March 16th 2009, 10:46 AM

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**Air** · March 16th 2009, 11:30 AM

Originally Posted by mr_motivator

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.

In this case, $y$ can be treated as a constant.

$\frac{\partial}{\partial x}\left(\frac{1}{7} \frac{\sin(6xy)}{x^4y}\right)$

Using product rule, we can set $u= \sin(6xy)\rightarrow u' = 6y\cos(6xy)$ and $v=\frac{1}{7yx^4}\rightarrow v'=\frac{-4}{7y}\left(\frac{1}{x^5}\right)$ .

Product rule: $\frac{\mathrm{d}(uv)}{\mathrm{d}x} = u'v+uv'$

$\therefore \frac{\partial}{\partial x}\left(\frac{1}{7} \frac{\sin(6xy)}{x^4y}\right) = \left(6y\cos(6xy)\right)\left(\frac{1}{7yx^4}\righ t) +\left(\sin(6xy)\right)\left(\frac{-4}{7y}\left(\frac{1}{x^5}\right)\right)$ $= \frac{6xy\cos(6xy) - 4\sin(6xy)}{7yx^5}$

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