## Calculus (3rd Edition)

$-2$
By making use of (FTC II), we have $$H'(x)=-\frac{d}{d x} \int_{9}^{4x^2} \frac{1}{t} dt =-\left(\frac{d}{d u} \int_{9}^{u} \frac{1}{t} dt\right)\frac{du}{dx}=-\frac{1}{4x^2} ( 8 x)\\ = -\frac{2}{x}.$$ Hence, $$H'(1)=-2.$$