Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 70

Determine the answer by using Chain Rule and check your answer by finding $f(g(x))$, taking the derivative and substituting.

$\displaystyle f(u) = 2u^5, g(x) = u = \frac{3-x}{4 + x}$

Find $(f \circ g)'(-10)$.



First, we find


$
\begin{equation}
\begin{aligned}

f'(u) = \frac{dy}{du} =& 10u^4
&&& g'(x) = \frac{du}{dx} =& \frac{\displaystyle (4 + x) \cdot \frac{d}{dx} (3-x) - (3-x) \cdot \frac{d}{dx} (4+x) }{(4+x)^2}
\\
\\
\phantom{=} & &&& =& \frac{(4+x) (-1) - (3-x) (1)}{(4+x)^2}
\\
\\
\phantom{=} & &&& =& \frac{-4-x - 3 + x}{(4+x)^2}
\\
\\
\phantom{=} & &&& =& \frac{-7}{(4+x)^2}

\end{aligned}
\end{equation}
$


Then,



$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} = 10u^4 \cdot \frac{-7}{(4+x)^2} =& \frac{-70u^4}{(4+x)^2}
\\
\\
=& \frac{\displaystyle -70 \left( \frac{3-x}{4+x} \right)^4 }{(4+x)^2}
\\
\\
=& \frac{-70 (3 - x)^4}{(4+x)^6}
\\
\\
(f \circ g)'(-10) =& \frac{-70 (3 + 10)^4}{(4-10)^6}
\\
\\
=& \frac{-70 (13)^4}{(-6)^6}
\\
\\
=& -42.85

\end{aligned}
\end{equation}
$


To check, we find first $f(g(x))$ and take the derivative.


$
\begin{equation}
\begin{aligned}

f(g(x)) = f \left( \frac{3 - x}{4 + x} \right) =& 2u^5
\\
\\
=& 2 \left( \frac{3-x}{4+x} \right)^5
\\
\\
f'(g(x)) =& 10 \left( \frac{3-x}{4+x} \right)^4 \cdot \frac{d}{dx} \left( \frac{3-x}{4+x} \right)
\\
\\
=& 10 \left( \frac{3-x}{4+x} \right)^4 \left[ \frac{\displaystyle (4+x) \cdot \frac{d}{dx} (3-x) - (3-x) \cdot \frac{d}{dx} (4+x) }{(4 + x)^2} \right]
\\
\\
=& 10 \left( \frac{3-x}{4+x} \right)^4 \left[ \frac{(4+x) (-1) - (3-x)(1)}{(4+x)^2} \right]
\\
\\
=& 10 \left( \frac{3-x}{4+x} \right)^4 \left[ \frac{-4 - x - 3 + x}{(4 + x)^2} \right]
\\
\\
=& \frac{-70 (3-x)^4}{(4+x)^6}
\\
\\
(f \circ g)' (-10) =& \frac{-70 [3 - (-10)]^4}{(4 - 10)^6}
\\
\\
=& \frac{-70 (13)^4}{(-6)^6}
\\
\\
=& -42.85



\end{aligned}
\end{equation}
$