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

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

2 Answers
Apr 3, 2018

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

Explanation:

We have:

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

By the chain rule

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!

Apr 3, 2018

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

Explanation:

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).