which function is increasing faster when x=1

Question

f(x)=(sqrt(x))^x^3 ------ g(x) = (cubicroot(x))^x^2

x>0

Answer 1

The rate of increase of a function, y(x), is just y'(x) or dy/dx.

Let's differentiate the 2 functions.

f(x) = (√x)^{x^3} [Take logs on both sides, whenever an exponent (power) involves a function of x, x^3

log f = x³ log (√x)

= x³ log (x^1/2) [Always replace roots with fractional powers for logs and differentiation

= x³ (1/2)log (x) [Now, we can differentiate, implicitly, on both sides

(1/f) f' = (3x²/2) log x + (1/2)x³ / x [Multiplication Rule

f' = f ( (3x²/2) log x + (1/2)x² )

= (√x)^{x^3} ( (3x²/2) log x + (1/2)x² ) [ Now to get f'(x) at x =1, plug 1 for x into this equation

At x = 1, f'(1) = (√1)^{1^3} ( (3(1)²/2) log x + (1/2)(1)² )

= 1 ( 0 + 1/2) [ Since log (1) = 0 in any base

= 3/2

g(x) = (cuberoot(x))^x^2 [Notice that the correct term is cube root, not cubic root

= (x^1/3)^{x^2} [Replacing the root by the correct fractional power

log g = x² log (x^1/3)

= x² (1/3) log (x) [By laws of logs

(1/g) g' = (2/3)x log x + (1/3) x² (1/x) [Multiplication Law

g' = g ( (2/3) x log x + (1/3) x ) [Multiply both sides by g

= (x^1/3)^{x^2} ( (2/3) x log x + (1/3) x ) [Replace g by its definition

=> g(1) = (1^1/3)^{1^2} ( (2/3) (1) log 1 + (1/3) 1 ) [Setting x =1 all the way through

= 1 ( (2/3) 0 + 1/3 )

= 1/3

Since 3/2 > 1/3, we know that function, f, is increasing faster than function, g.

Answer 2

Hi Okabe R.,

It's helpful to first simplify these functions before differentiating.

f(x)=x^0.5^x^3 and g(x)=x^(1/3)^x^2; since a^b^c = a^(b*c) these reduce to:

f(x)=x^(1.5*x) and g(x)=x^(0.66...*x) ; then convert to powers of e, in order to set up Chain Rule

f(x)=e^ln(x)^(1.5*x) and g(x)=e^ln(x)^0.66...*x)

So f '(x)=e^(ln(x)*1.5*x) *(1.5)*(x^-1*x + ln(x)) and g ' (x) = e^(ln(x)*0.66...*x)*(0.66...)*(x^-1*x + ln(x))

f ' (x) = 1^1.5 * 1.5 * 1 and g ' (x) = 1^0.66 * 0.66 * 1 since 1^n = 1 for any n

f ' (1) = 1.5 ; g '(1) = 0.66...

-- Cheers, -- Mr. d.