What is the derivative of #sqrt(4+x^2) #?

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Answer 1 · 2016-04-02T15:09:36

#dy/dx = x/sqrt(4+x^2)#

Change the function #f(x) = sqrt(4 + x^2)# into something more manageable by the laws of indices: #sqrt(u) = u^(1/2)#

#f(x) = (4 + x^2)^(1/2)#

This is approachable with the chain rule, which states that if #f(x) = g(h(x))# then #f'(x) = h'(x)g'(h)#, where #h# is just shorthand for #h(x)#.

With #h(x) = 4 + x^2# and #g(x) = h^2(x)# and simple differentiation,

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

Multiplying these two together as #h'(x)g'(h)#,

#(2x)/(2sqrt(4 + x^2)) = x/sqrt(4+x^2)#