Take the derivative of this:...?

Question

Take the derivative of this, and show all steps:

f(x) = x sqrt(4 - x^2)

Thanks :)

Answer 1

f(x) = y = x sqrt(4 -x^2)

y = x (4 -x^2)^1/2

y ' = (x)' (4-x^2)^1/2 + ((4-x^2)^1/2)' (x)

y ' = (1) (4-x^2)^1/2 + (1/2(4-x^2)^-1/2) ((4)' + (-x^2)') (x)

y ' = (4-x^2)^1/2 + 1/2(4-x^2)^-1/2 ( -2x^2)

y ' = sqrt(4-x^2) + ( - 2x^2)/sqrt(4-x^2))

y ' = sqrt(4-x^2) -x^2/sqrt(4-x^2)

Answer 2

Hello,

The derivative of the first times the second + the derivative of the second times the first.

f'(x) = 1 * sqrt(4 - x^2) + 1/2(4 - x^2)^(-1/2) * (-2x) =

f'(x) = sqrt(4 - x^2) + x/(2*(4-x^2)^1/2) Now you can simply this.

Hope This Helps!

Answer 3

Use the product rule:
d/dx ( f(x)*g(x))=f'(x)g(x) + f(x)g'(x)
where in your case,
f(x)=x, f'(x)=1,
g(x)=sqrt(4-x^2),
g'(x)=(1/2)*(4-x^2)^(-0.5) *(-2x),
g'(x)=-x/sqrt(4-x^2)
where g'(x) is obtained via the chain rule.

so the derivative is
(-x/sqrt(4-x^2))+x*sqrt(4-x^2)
=x*(sqrt(4-x^2) - (1/sqrt(4-x^2)))

Answer 4

y = xsqrt(4 - x^2) = x * (4 - x^2)^(1/2)
y' = sqrt(4 - x^2) + x * (-2x) * 1/2 * (4 - x^2)^(-1/2)
y' = sqrt(4 - x^2) - x^2 / sqrt(4 - x^2)

Edit: Derivative of u x v = u'. v + u . v'

Answer 5

Use the chain rule and product rule:
f ' (x) = [ 1*√(4 - x^2) ] + [x * (1/2)(4 - x^2)^(-1/2) * (-2x) ]

Now simplify