Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 38

Differentiate $\displaystyle h(x) = \left( \frac{1-3x}{2-7x} \right)^{-5}$.



$
\begin{equation}
\begin{aligned}

h(x) =& \left( \frac{2 - 7x}{1 - 3x} \right)^5
\qquad \text{Laws of Exponent}
\\
\\
h'(x) =& 5 \left( \frac{2-7x}{1-3x} \right)^4 \cdot \frac{d}{dx} \left( \frac{2-7x}{1-3x} \right)
\\
\\
h'(x) =& 5 \left( \frac{2-7x}{1-3x} \right)^4 \left[ \frac{\displaystyle (1-3x) \cdot \frac{d}{dx} (2-7x) - (2-7x) \cdot \frac{d}{dx} (1-3x) }{(1-3x)^2} \right]
\\
\\
h'(x) =& 5 \left( \frac{2-7x}{1-3x} \right)^4 \left[ \frac{(1 - 3x) (-7) - (2-7x) (-3)}{(1-3x)^2} \right]
\\
\\
h'(x) =& 5 \left( \frac{2-7x}{1-3x} \right)^4 \left[ \frac{-7+21x + 6 -21x}{(1 - 3x)^2} \right]
\\
\\
h'(x) =& 5 \left( \frac{2-7x}{1-3x} \right)^4 \left[ \frac{-1}{(1 - 3x)^2} \right]
\\
\\
h'(x) =& \frac{-5(2 - 7x)^4}{(1-3x)^6}

\end{aligned}
\end{equation}
$