Find g'(4) and compute the limit

1. (a) Compute g ' (4), given g(x) = integral (from t = 1 to sqrt(x))
ln(t^(2) + 1)dt.


(b) Compute lim x-->0 [5^x - 3^x]/[ln(1 + x)].

1) Using the Fundamental Theorem of Calculus with the Chain Rule,
g'(x) = ln((√x)^2 + 1) * (d/dx) √x
.......= ln(x + 1) * (1/2)x^(-1/2).

Letting x = 4 yields g'(4) = (1/4) ln 5.
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2) lim(x→0) (5^x - 3^x) / ln(1 + x); this is of the form "0/0"
= lim(x→0) (5^x ln 5 - 3^x ln 3) / (1/(1 + x)), by L'Hopital's Rule
= ln 5 - ln 3
= ln(5/3).

I hope this helps!

(a)

Let G(t) = ∫ ln(t²+1) dt -----> G'(t) = ln(t²+1)

g(x) = ∫[1 to √x] ln(t²+1) dt
g(x) = G(√x) - G(1)

g'(x) = G'(√x) * d/dx (√x) - 0
g'(x) = ln(x+1) * 1/(2√x)

g'(4) = ln(4+1) * 1/(2√4) = ln(5) / 4

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(b)

lim[x→0] (5^x - 3^x) / (ln(1+x)) = (1-1)/ln(1+0) = 0/0

Use L'Hopital's rule:
= lim[x→0] (ln(5) 5^x - ln(3) 3^x) / [1/(1+x)]
= (ln(5) - ln(3)) / (1/1)
= ln(5) - ln(3)
= ln(5/3)

Mαthmφm