Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 62

Suppose that $h(x) = \sqrt{4 + 3 f(x)}$, where $f(1) = 7$ and $f'(1) = 4$, find $h'(1)$.


$
\begin{equation}
\begin{aligned}

h'(x) =& \frac{d}{dx} \left(\sqrt{4 + 3 f(x)} \right)
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\\
h'(x) =& \frac{d}{dx} (4 + 3 f(x))^{\frac{1}{2}}
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h'(x) =& \frac{1}{2} (4 + 3 f(x))^{\frac{-1}{2}} \cdot \frac{d}{dx} (4 + 3 f(x))
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h'(x) =& \frac{1}{2} (4 + 3 f(x))^{\frac{-1}{2}} \left[ \frac{d}{dx} (4) + 3 \frac{d}{dx} (f(x)) \right]
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h'(x) =& \frac{1}{2} (4 + 3 f(x))^{\frac{-1}{2}} (0 + 3 f'(x))
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h'(x) =& \frac{1}{2} (4 + 3 f(x))^{\frac{-1}{2}} (3 f'(x))
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h'(x) =& \frac{3 f'(x)}{2 \sqrt{4 + 3 f(x)}}
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h'(1) =& \frac{3 f'(1)}{2 \sqrt{4 + 3 f(1)}}
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h'(1) =& \frac{(3)(4)}{2 \sqrt{ 4 + (3) (7)}}
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h'(1) =& \frac{12}{2 \sqrt{4 + 21}}
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h'(1) =& \frac{6}{\sqrt{25}}
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h'(1) =& \frac{6}{5}

\end{aligned}
\end{equation}
$