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

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Answer 1 · 2016-12-19T13:41:50

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)#