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

2 Answers
Jan 21, 2016

Answer:

#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

Answer:

#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)#