# Question bbb77

##### 2 Answers
Jan 15, 2017

$y = \sin \left(\frac{\pi}{6} {e}^{x y}\right)$

Differentiating w.r to x we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\frac{\pi}{6} {e}^{x y}\right) \times \frac{\pi}{6} {e}^{x y} \times \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\frac{\pi}{6} {e}^{x y}\right) \times \frac{\pi}{6} {e}^{x y} \times \left(\sin \left(\frac{\pi}{6} {e}^{x y}\right) + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Putting $x = 0$

((dy)/(dx))_(x=0)=cos(pi/6e^(0*y))xxpi/6e^(0*y)xx(sin(pi/6)e^(0*y))+0*(dy)/(dx))#

$= \cos \left(\frac{\pi}{6}\right) \times \frac{\pi}{6} \times \sin \left(\frac{\pi}{6}\right)$

$= \frac{\sqrt{3}}{2} \times \frac{\pi}{6} \times \frac{1}{2}$

$= \frac{\sqrt{3} \pi}{24}$

Jan 15, 2017

$y ' \left(0\right) = \frac{\pi \sqrt{3}}{24}$

#### Explanation:

We have the equation:

$y = \sin \left(\frac{\pi}{6} {e}^{x y}\right)$

differentiate with respect to $x$:

$y ' = \left(y + x y '\right) \frac{\pi}{6} {e}^{x y} \cos \left(\frac{\pi}{6} {e}^{x y}\right)$

Solving for $y '$:

$y ' = \frac{\frac{\pi}{6} y {e}^{x y} \cos \left(\frac{\pi}{6} {e}^{x y}\right)}{1 - \frac{\pi}{6} x {e}^{x y} \cos \left(\frac{\pi}{6} {e}^{x y}\right)}$

Now, for $x = 0$ we have:

$y \left(0\right) = \sin \left(\left(\frac{\pi}{6}\right) {e}^{0 \cdot y}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

and then:

$y ' \left(0\right) = \frac{\frac{\pi}{6} \frac{1}{2} \cos \left(\frac{\pi}{6}\right)}{1} = \frac{\pi \sqrt{3}}{24}$