What is the first differential of $y = e^{sin \sqrt{x}}$ $y= e^sinsqrtx$ ?

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Answer 1 · 2017-01-07T05:22:25

$\frac{d y}{d x} = \frac{e^{sin (\sqrt{x})} \cdot cos (\sqrt{x})}{2 \sqrt{x}}$ $dy/dx=(e^(sin(sqrtx))*cos(sqrtx))/(2sqrtx)$

$y = e^{sin \sqrt{x}}$ $y= e^sinsqrtx$

Applying the chain rule:

$\frac{d y}{d x} = e^{sin (\sqrt{x})} \cdot \frac{d}{d x} (sin (\sqrt{x}))$ $dy/dx=e^(sin(sqrtx)) * d/dx(sin(sqrtx))$

Applying the chain rule again:

$\frac{d y}{d x} = e^{sin (\sqrt{x})} \cdot cos (\sqrt{x}) \cdot \frac{d}{d x} (\sqrt{x})$ $dy/dx=e^(sin(sqrtx)) * cos(sqrtx) * d/dx(sqrtx)$

Applying the power rule:

$\frac{d y}{d x} = e^{sin (\sqrt{x})} \cdot cos (\sqrt{x}) \cdot \frac{1}{2} x^{- \frac{1}{2}}$ $dy/dx=e^(sin(sqrtx)) * cos(sqrtx) * 1/2x^(-1/2)$

$\frac{d y}{d x} = \frac{e^{sin (\sqrt{x})} \cdot cos (\sqrt{x})}{2 \sqrt{x}}$ $dy/dx=(e^(sin(sqrtx))*cos(sqrtx))/(2sqrtx)$

What is the first differential of y= e^sinsqrtxy=esin√xy= e^sinsqrtx ?