Question

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

Answer 1

`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))`

Answer 2

`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

Answer 3

`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)`