2 quick question on 2nd ftc

[kpx001]

give g(x) = [4x,1] ln(t^2+t)dt, find d(g(x))/dx

basiclly i thought the answer was 4ln(16x^2+4x) but the answer is -4ln(16x^2+4x). does it have to do with the 4x being a, instead of b which makes it either negative or positive? i would assume so because f(b)-f(a) => -f(a) ? also is doing 2nd ftc basicly like finding f'' ?

 

[skeeter]

2nd FTC ...

\(\displaystyle \frac{d}{dx} \left[\int_a^u f(t) \, dt \right] = f(u) \cdot \frac{du}{dx}\)

your function ...

\(\displaystyle g(x) = \int_{4x}^1 \ln(t^2 + t) \, dt = -\int_1^{4x} \ln(t^2 + t) \, dt\)

\(\displaystyle \frac{d}{dx} \left[-\int_1^{4x} \ln(t^2 + t) \, dt\right] = -4\ln(16x^2 + 4x)\)