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

2 Answers
Jan 21, 2016

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

Jan 21, 2016

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)