How do you differentiate #g(x) =(2x^2+1)^(3/2) (x-1)^(7/4) # using the product rule?

1 Answer
Apr 4, 2016

Answer:

#6xsqrt(2x^2 + 1)root(4)(x-1)^7 + 7/4root(4)(x-1)^3root(2)(2x^2+1)^3#

Explanation:

The product rule states that if

#g(x) = s(x)*t(x)#

then

#g'(x) = s'(x)t(x) + t'(x)s(x)#.

If we substitute #s(x) = (2x^2 + 1)^(3/2)# and #t(x) = (x-1)^(7/4)#, we can find #s'(x)# and #t'(x)# using the chain rule, because each function is one function (in the brackets) inside another (the index or power).

The chain rule says that if

#h(x) = k(l(x)) -> h'(x) = l'(x)k'(l)#

Therefore,

#d/dx (2x^2 + 1) = 4x#
#d/dx u^(3/2) = 3/2u^(1/2)#, where #u# can be #(2x^2 + 1)#.
#s'(x) = 4x * 3/2(2x^2 + 1)^(1/2) = 6xsqrt(2x^2 + 1)#

and for #t(x)#,

#d/dx (x-1) = 1#
#d/dx u^(7/4) = 7/4u^(3/4)#
#t'(x) = 7/4(x-1)^(3/4) = 7/4root(4)(x-1)^3#

Therefore, by substituting into the product rule,

#s'(x)t(x) + t'(x)s(x) = #
#6xsqrt(2x^2 + 1)root(4)(x-1)^7 + 7/4root(4)(x-1)^3root(2)(2x^2+1)^3#