# Differentiate with respect to y : (2x-cos3y)^4sec(ln(1-xy))+√(1+(√y))?

Aug 17, 2015

$\sec \left(\ln \left(1 - x y\right)\right) \left(4 {\left(2 x - \cos 3 y\right)}^{3} \left(2 \frac{\mathrm{dx}}{\mathrm{dy}} + 3 \sin 3 y\right)\right) + {\left(2 x - \cos 3 y\right)}^{4} \left(\sec \left(\ln \left(1 - x y\right)\right) \tan \left(\ln \left(1 - x y\right)\right) \cdot \frac{\left(y \frac{\mathrm{dx}}{\mathrm{dy}} + x\right)}{x y - 1}\right) + \frac{1}{4 \sqrt{y + y \sqrt{y}}}$

#### Explanation:

It's an atrocious chain rule phenomenon

Let ${\left(2 x - \cos 3 y\right)}^{4} \sec \left(\ln \left(1 - x y\right)\right) + \sqrt{1 + \sqrt{y}} = p q + r$
Where $p = {\left(2 x - \cos 3 y\right)}^{4} , q = \sec \left(\ln \left(1 - x y\right)\right) , r = \sqrt{1 + \sqrt{y}}$

$\frac{\mathrm{dp}}{\mathrm{dy}} = 4 {\left(2 x - \cos 3 y\right)}^{3} \left(2 \frac{\mathrm{dx}}{\mathrm{dy}} + 3 \sin 3 y\right)$

$\frac{\mathrm{dq}}{\mathrm{dy}} = \sec \left(\ln \left(1 - x y\right)\right) \tan \left(\ln \left(1 - x y\right)\right) \cdot \left(\frac{- \left(\frac{\mathrm{dx}}{\mathrm{dy}} y + x\right)}{1 - x y}\right)$

$\frac{\mathrm{dq}}{\mathrm{dy}} = \sec \left(\ln \left(1 - x y\right)\right) \tan \left(\ln \left(1 - x y\right)\right) \cdot \frac{\left(y \frac{\mathrm{dx}}{\mathrm{dy}} + x\right)}{x y - 1}$

$\frac{\mathrm{dr}}{\mathrm{dy}} = \frac{\frac{1}{2 \sqrt{y}}}{2 \sqrt{1 + \sqrt{y}}} = \frac{1}{4 \sqrt{y + y \sqrt{y}}}$

Thus $\frac{d}{\mathrm{dy}} {\left(2 x - \cos 3 y\right)}^{4} \sec \left(\ln \left(1 - x y\right)\right) + \sqrt{1 + \sqrt{y}}$
$= \frac{d}{\mathrm{dy}} \left(p q + r\right)$
$= \frac{d}{\mathrm{dy}} \left(p q\right) + \frac{d}{\mathrm{dy}} r$
$= q \left(\frac{\mathrm{dp}}{\mathrm{dy}}\right) + p \left(\frac{\mathrm{dq}}{\mathrm{dy}}\right) + \frac{\mathrm{dr}}{\mathrm{dy}}$
$= \sec \left(\ln \left(1 - x y\right)\right) \left(4 {\left(2 x - \cos 3 y\right)}^{3} \left(2 \frac{\mathrm{dx}}{\mathrm{dy}} + 3 \sin 3 y\right)\right) + {\left(2 x - \cos 3 y\right)}^{4} \left(\sec \left(\ln \left(1 - x y\right)\right) \tan \left(\ln \left(1 - x y\right)\right) \cdot \frac{\left(y \frac{\mathrm{dx}}{\mathrm{dy}} + x\right)}{x y - 1}\right) + \frac{1}{4 \sqrt{y + y \sqrt{y}}}$