# How do you differentiate  f(t)=-e^(sin(pi/x))sinpix  using the chain rule.?

Apr 10, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\pi {e}^{\sin \left(\frac{\pi}{x}\right)} \sin \left(\pi x\right) \cos \left(\frac{\pi}{x}\right)}{{x}^{2}} - \pi {e}^{\sin \left(\frac{\pi}{x}\right)} \cos \left(\pi x\right)$

#### Explanation:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)} \sin \left(\pi x\right)\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)}\right] \sin \left(\pi x\right) + \frac{d}{\mathrm{dx}} \left[\sin \left(\pi x\right)\right] \left(- {e}^{\sin \left(\frac{\pi}{x}\right)}\right)$

$\frac{d}{\mathrm{dx}} \left[\sin \left(\pi x\right)\right] \left(- {e}^{\sin \left(\frac{\pi}{x}\right)}\right) = \pi \cos \left(\pi x\right) \left(- {e}^{\sin \left(\frac{\pi}{x}\right)}\right)$

$\frac{d}{\mathrm{dx}} \left[\sin \left(\pi x\right)\right] \left(- {e}^{\sin \left(\frac{\pi}{x}\right)}\right) = - \pi {e}^{\sin \left(\frac{\pi}{x}\right)} \cos \left(\pi x\right)$

Chain rule time: $\alpha = \sin \left(\frac{\pi}{x}\right) , \beta = \frac{1}{x}$

$\frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)}\right] = \frac{d}{\mathrm{da} l p h a} \left[- {e}^{\alpha}\right] \frac{d}{\mathrm{db} \eta} \left[\sin \left(\pi \beta\right)\right] \frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]$

$\frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)}\right] = - {e}^{\alpha} \times \pi \cos \left(\pi \beta\right) \times \left(- \frac{1}{x} ^ 2\right)$

$\frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)}\right] = - {e}^{\sin \left(\frac{\pi}{x}\right)} \times \pi \cos \left(\frac{\pi}{x}\right) \times \left(- \frac{1}{x} ^ 2\right)$

$\frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)}\right] = \frac{\pi {e}^{\sin \left(\frac{\pi}{x}\right)} \cos \left(\frac{\pi}{x}\right)}{{x}^{2}}$

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$\frac{d}{\mathrm{dx}} \left[- {e}^{\sin \left(\frac{\pi}{x}\right)}\right] \sin \left(\pi x\right) = \frac{\pi {e}^{\sin \left(\frac{\pi}{x}\right)} \sin \left(\pi x\right) \cos \left(\frac{\pi}{x}\right)}{{x}^{2}}$

Adding the two together we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\pi {e}^{\sin \left(\frac{\pi}{x}\right)} \sin \left(\pi x\right) \cos \left(\frac{\pi}{x}\right)}{{x}^{2}} - \pi {e}^{\sin \left(\frac{\pi}{x}\right)} \cos \left(\pi x\right)$