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What is the first differential of #y= e^sinsqrtx# ?

Answer 1

Aurora Alvarado

#dy/dx=(e^(sin(sqrtx))*cos(sqrtx))/(2sqrtx)#

#y= e^sinsqrtx#

Applying the chain rule:

#dy/dx=e^(sin(sqrtx)) * d/dx(sin(sqrtx))#

Applying the chain rule again:

#dy/dx=e^(sin(sqrtx)) * cos(sqrtx) * d/dx(sqrtx)#

Applying the power rule:

#dy/dx=e^(sin(sqrtx)) * cos(sqrtx) * 1/2x^(-1/2)#

#dy/dx=(e^(sin(sqrtx))*cos(sqrtx))/(2sqrtx)#

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Answer 2

Christopher Agresta

The first differential of ( y = e^{\sin(\sqrt{x})} ) is:

[ \frac{dy}{dx} = \frac{d}{dx} \left( e^{\sin(\sqrt{x})} \right) = e^{\sin(\sqrt{x})} \cdot \frac{d}{dx} \left( \sin(\sqrt{x}) \right) ]

Using the chain rule and the derivative of (\sin(\sqrt{x})), we get:

[ \frac{dy}{dx} = e^{\sin(\sqrt{x})} \cdot \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.