Question

How do you differentiate `y =x sqrt((4-x^2)` using the chain rule?

Answer 1

`h'(x) = sqrt(4-x^2) - 1/(sqrt(4-x^2)) x^2`
`h'(x) = - (2x^2-4)/ (sqrt(4-x^2))`

Explanation:

This actually a product of 2 functions:
Let
Use the product rule and write:

1) `h(x) = f(x)g(x) and h'(x)= f'(x)g(x) + f(x)g'(x)`

We use the chain rule on g(x):
Let `g(x) = (r@p)(x)`
then
2) `g'(x) = r'(p(x))*p'(x)`
Now: `r(x) =sqrt(x); p(x) = 4-x^2`
`r'(x) = 1/(2sqrt(x)); p'(x) = -2x`
`g'(x) = 1/(cancel(2)sqrt(4-x^2)) *cancel((-2)x)`

`g'(x) = -1/(sqrt(4-x^2)) *(x)`

So what we have is:
`f(x) = x; f'(x)= 1`
`g(x) = sqrt(4-x^2); g'(x) =-1/(sqrt(4-x^2)) x`

From 1)
`h'(x) = sqrt(4-x^2) - 1/(sqrt(4-x^2)) x^2`

Answer 2

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

Explanation:

`y=f(x)*g(x)`

`f(x)=x`
`g(x)=h(i(x))=sqrt(i(x))`
`i(x)=(4-x^2)`

we have to use Product Rule and Chain Rule:

`y'=f'(x)*g(x)+f(x)g'(x)`

`g'(x)=d/dxh(i(x))=h'(i(x))*i'(x)`

therefore:

`y'=1*sqrt((4-x^2))+x*(1/(2sqrt((4-x^2)))*(0-2x))=`

`=sqrt((4-x^2))+x*(1/(cancel(2)sqrt((4-x^2)))(-cancel(2)x))=`
`=sqrt((4-x^2))-x^2/sqrt((4-x^2))`
`=(4-x^2-x^2)/sqrt(4-x^2)=2*(2-x^2)/sqrt(4-x^2)`