Derivative Problem - # 2

[Jason76]

\(\displaystyle g(x) = \int_{x}^{6} \cos \sqrt{7t} dt\)

What is \(\displaystyle g'(x)\)?

The homework says the answer is \(\displaystyle -\cos \sqrt{7x}\) But shouldn't it be \(\displaystyle \cos \sqrt{7x}\)

 

[Romsek]

\(\displaystyle g(x) = \int_{x}^{6} \cos \sqrt{7t} dt\)

What is \(\displaystyle g'(x)\)?

The homework says the answer is \(\displaystyle -\cos \sqrt{7x}\) But shouldn't it be \(\displaystyle \cos \sqrt{7x}\)

let F(t) be the antiderivative of f(t) = cos(sqrt(7t))

g(x) = Integral[x,6,f(t)dt]

g(x) = F(6) - F(x)

g'(x) = 0 - F'(x)

g'(x) = -f(x)

you need the x in the upper limit to avoid the negative sign.