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

Answer 1

Emily Ammerman

The answer is #=-x/(16-x^2)#

This is a chain differentiation

#y=ln( sqrtu(x))#

#y=1/2*ln(u(x))#

#dy/dx=1/2*1/(u(x))*u'(x)#

#y=ln(sqrt(16-x^2))#

#y=1/2*ln(16-x^2)#

Therefore,

#dy/dx=1/2*1/(16-x^2)*(-2x)#

#dy/dx=-x/(16-x^2)#

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Answer 2

Paisley Johnson

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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Answer from HIX Tutor

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