What is the derivative of #f(x) = x(sqrt( 1 - x^2))#?

Answer 1

Isaiah Allen

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

We will require the use of two rules: the product rule and the chain rule. The product rule states that:

#(d(fg))/dx# = #(df)/dx * g(x) + f(x) * (dg)/dx#.

The chain rule states that:

#(dy)/dx = (dy)/(du) (du)/dx#, where #u# is a function of #x# and #y# is a function of #u#.

Therefore,

#(df)/dx = (x)' * (sqrt(1-x^2)) + x * (sqrt(1-x^2))'#

To find the derivative of #sqrt(1-x^2)#, use the chain rule, with

#u = 1-x^2: (sqrtu)' = 1/(2sqrtu) * u'#

#= -(2x)/(2(sqrt(1-x^2))# #= -x/(sqrt(1-x^2))#.

Substituting this result into the original equation:

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

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isabella Ackerman

The derivative of ( f(x) = x\sqrt{1 - x^2} ) is ( f'(x) = \sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.