Answer
$h(\ln x)=4.3(\ln x)^3-2 (\ln x)^2+4 (\ln x)-12$
$\frac{dh( \ln x)}{dx}=12.9(\ln x)^2 \frac{1}{x} -4(\ln x) \frac{1}{x}+4\cdot \frac{1}{x}$
Work Step by Step
$h(t)=4.3t^3-2t^2+4t-12$
$t(x)=\ln x$
Composite function will be
$y=h(\ln x)=4.3(\ln x)^3-2 (\ln x)^2+4 (\ln x)-12$
Taking derivative with respect to x
$\frac{dy}{dx}=4.3\times3 (\ln x)^2 \frac{1}{x} -2\times2 (\ln x) \frac{1}{x}+4\times \frac{1}{x}$
$\frac{d{y}}{dx}=12.9(\ln x)^2 \frac{1}{x} -4 (\ln x) \frac{1}{x}+4\cdot \frac{1}{x}$
