# What is the derivative of #y= (sqrt x) ^x#?

We can write the function as:

Now:

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Differentiate the left hand side using implicit differentiation and the right hand using the chain and product rules. Before using the product rule, we must use the chain rule.

Now to the rest of the function:

Hopefully this helps!

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To find the derivative of ( y = (\sqrt{x})^x ), you can use logarithmic differentiation. The steps are as follows:

- Take the natural logarithm of both sides of the equation.
- Apply the logarithmic properties to simplify the expression.
- Differentiate both sides with respect to ( x ).
- Solve for ( \frac{{dy}}{{dx}} ).

Starting with ( y = (\sqrt{x})^x ):

[ \ln(y) = \ln\left((\sqrt{x})^x\right) ]

[ \ln(y) = x \ln(\sqrt{x}) ]

Now, differentiate both sides with respect to ( x ):

[ \frac{1}{y} \frac{dy}{dx} = \ln(\sqrt{x}) + \frac{x}{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} ]

[ \frac{1}{y} \frac{dy}{dx} = \ln(\sqrt{x}) + \frac{x}{2x} ]

[ \frac{1}{y} \frac{dy}{dx} = \ln(\sqrt{x}) + \frac{1}{2} ]

Now, solve for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = y \left( \ln(\sqrt{x}) + \frac{1}{2} \right) ]

[ \frac{dy}{dx} = (\sqrt{x})^x \left( \ln(\sqrt{x}) + \frac{1}{2} \right) ]

Thus, the derivative of ( y = (\sqrt{x})^x ) is ( \frac{dy}{dx} = (\sqrt{x})^x \left( \ln(\sqrt{x}) + \frac{1}{2} \right) ).

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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.

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.

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.

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