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

1 Answer
Aug 19, 2015

Answer:

#g^' = (x * (2 - 3x^2))/(sqrt(1-x^2))#

Explanation:

You can use the product rule and the chain rule to differentiate this function, which can be rewritten as

#g(x) = x^2 * (1-x^2)^(1/2)#

The product rule allows you to differentiate functions that take the form

#color(blue)(g(x) = f(x) * h(x))#

by using the formula

#color(blue)(d/dx(g(x)) = [d/dx(f(x))] * h(x) + f(x) * d/dx(h(x))#

In your case, you have

#f(x) = x^2" "# and #" "h(x) = (1-x^2)^(1/2)#

This means that you can write

#d/dx(g(x)) = [d/dx(x^2)] * (1-x^2)^(1/2) + x^2 * d/dx(1-x^2)^(1/2)#

To find #d/dx(h(x))#, you need to use the chain rule for #u^(1/2)#, with #u = (1-x^2)#. This will get you

#d/dx(u^(1/2)) = d/(du)u^(1/2) * d/dx(u)#

#d/dx(u^(1/2)) = 1/2 * u^(-1/2) * d/dx(1-x^2)#

#d/dx(1-x^2)^(1/2) = 1/color(red)(cancel(color(black)(2))) * (1-x^2)^(-1/2) * (-color(red)(cancel(color(black)(2)))x)#

#d/dx(1-x^2)^(1/2) = -x * (1-x^2)^(-1/2)#

Plug this into your target derivative to get

#g^' = 2x * (1-x^2)^(1/2) + x^2 * [-x * (1-x^2)^(-1/2)]#

This can be simplified to get

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

#g^' = x * (1-x^2)^(-1/2) * (2 - 3x^2)#

Finally, you can rewrite this as

#g^' = color(green)((x * (2 - 3x^2))/(sqrt(1-x^2)))#