What is the value of #f^'(2)#?

Question

Given, #f(x)=int_1^(x^3)#(1/(1+lnt))dt

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Answer 1 · 2018-04-03T18:57:32

#f'(2) = 12/(1 +ln8)#

We have:

#f(x) = int_1^(x^3) 1/(1 +lnt) dt#

#f'(x) = (3x^2)/(1 + ln(x^3)) #

Thus

#f'(2) = (3(2)^2)/(1 + ln(8))#

#f'(2) = 12/(1 +ln8)#

Hopefully this helps!

Answer 2 · 2018-04-03T19:15:41

# 12/(1+ln8)=12/ln(8e)#.

Suppose that, #int1/(1+lnt)dt=g(t)+C#.

#:. g'(t)=1/(1+lnt)...[" the Definition of Integral]"...(ast)#.

Further, by the Fundamental Theorem of Calculus,

#f(x)=int_1^(x^3)1/(1+lnt)dt=[g(t)+C]_1^(x^3)=g(x^3)-g(1)#.

# rArr f'(x)=d/dx{g(x^3)-g(1)}#,

#=g'(x^3)*d/dx(x^3)-0......[because," The Chain Rule]"#.

#=3x^2*g'(x^3)#,

#=3x^2*1/(1+lnx^3)............[because, (ast)]#,

#:. f'(x)=(3x^2)/(1+lnx^3)#.

# rArr f'(2)=(3*2^2)/(1+ln2^3)=12/(lne+ln8)=12/ln(8e)#.