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

Question

Elijah Abram · Accepted Answer

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

Christopher Agresta · Answer

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

#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)#