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

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

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)