How do you find the derivative of #y =sqrt(1-x^2)#?

3 Answers
Jul 29, 2014

#y'=-x/(sqrt(1-x^2))#

Using Chain Rule

#y=sqrt(f(x))#

#y'=1/(2sqrt(f(x)))*f'(x)#

Similarly, following for the above function,

#y'=(sqrt(1-x^2))'#

#y'=1/(2sqrt(1-x^2))*(1-x^2)'#

#y'=1/(2sqrt(1-x^2))*(-2x)#

#y'=-x/(sqrt(1-x^2))#

Jun 3, 2017

Answer:

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

Explanation:

Another method is using implicit differentiation like this:

#y = sqrt(1-x^2)#

#y^2 = 1 - x^2#

#x^2 + y^2 = 1#

Now differentiate with respect to #x#:

#2x + 2y(dy/dx) = 0#

#x + y(dy/dx) = 0#

#y(dy/dx) = -x#

#(dy/dx) = -x/y#

Finally, make the substitution #y = sqrt(1-x^2)#

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

Final Answer

Jun 3, 2017

Answer:

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

Explanation:

#y=sqrt(1-x^2)#

Now let #u=1-x^2#

#y=u^(1/2)#

#dy/(du)=1/2(u^(-1/2))=1/(2u^(1/2))=1/(2sqrtu)=1/(2sqrt(1-x^2))#

#(du)/dx=-2x#

Now apply the chain rule:

#dy/(cancel(du))xx(cancel(du))/dx=dy/dx#

#dy/dx=-2x xx1/(2sqrt(1-x^2))=-x/sqrt(1-x^2)#