How to het f'(x) using the quotient rule on this function: 2x-3 divided by x^2.5?

Question

#(2x-3)/(2x^2.5)#

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Answer 1 · 2018-06-12T19:08:31

#f'(x)=-3/2*x^(-5/2)+15/4*x^(-7/2)#

Writing your #f(x)# in the form
#f(x)=(2x)/(2*x^(5/2))-3/(2*x^(5/2))#
and this is

#f(x)=x^(-3/2)-3/2*x^(-5/2)#
so
#f'(x)=-3/2x^(-5/2)+15/4*x^(-7/2)#

Answer 2 · 2018-06-13T01:03:50

#f'(x) = frac{15sqrtx - 6x^(3/2)}{4x^4}#

#f(x) = frac{2x-3}{2x^(5/2)}#

Note that the quotient rule states that if #color(blue)(f(x) = g(x)/(h(x)))#, then #color(blue)(f'(x) = frac{h(x)g'(x)-g(x)h'(x)}{(h(x))^2})#.

We apply the quotient rule to this context:

#f'(x) = frac{(2x^(5/2))(2)-(2x-3)(5/2 * 2 * x^(3/2))}{(2x^(5/2))^2}#

Simplify:
#f'(x) = frac{4x^(5/2) - (2x-3)(5x^(3/2))}{4x^5}#

Simplify further by distributing:
#f'(x) = frac{4x^(5/2) - 10x^(5/2) + 15x^(3/2)}{4x^5}#

#f'(x) = frac{-6x^(5/2) + 15x^(3/2)}{4x^5}#

#f'(x) = frac{x(-6x^(3/2) + 15x^(1/2))}{4x^5}#

#f'(x) = frac{15sqrtx - 6x^(3/2)}{4x^4}#