# If #y=ln(sqrt(16-x^2))#, then what is #dy/dx#?

The answer is

This is a chain differentiation

Therefore,

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To find ( \frac{dy}{dx} ) for ( y = \ln(\sqrt{16 - x^2}) ), you can use the chain rule and the derivative of the natural logarithm function. The derivative is:

[ \frac{dy}{dx} = \frac{1}{\sqrt{16 - x^2}} \times \frac{d}{dx}(\sqrt{16 - x^2}) ]

Now, differentiate ( \sqrt{16 - x^2} ) with respect to ( x ):

[ \frac{d}{dx}(\sqrt{16 - x^2}) = \frac{1}{2\sqrt{16 - x^2}} \times (-2x) ]

Simplify:

[ \frac{d}{dx}(\sqrt{16 - x^2}) = \frac{-x}{\sqrt{16 - x^2}} ]

Substitute this back into the expression for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \frac{1}{\sqrt{16 - x^2}} \times \left(\frac{-x}{\sqrt{16 - x^2}}\right) ]

[ \frac{dy}{dx} = \frac{-x}{(16 - x^2)} ]

So, ( \frac{dy}{dx} = \frac{-x}{(16 - x^2)} ).

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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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