To find [tex]\( h'(4) \)[/tex] given [tex]\( h(x) = \sqrt{4 + 3f(x)} \)[/tex], where [tex]\( f(4) = 4 \)[/tex] and [tex]\( f'(4) = 3 \)[/tex], let's perform the following steps:
1. Differentiate [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
Given [tex]\( h(x) = \sqrt{4 + 3f(x)} \)[/tex], we need to differentiate this with respect to [tex]\( x \)[/tex].
First, let's set:
[tex]\[
u = 4 + 3f(x)
\][/tex]
Thus, [tex]\( h(x) = \sqrt{u} \)[/tex].
The derivative [tex]\( h'(x) \)[/tex] can be found using the chain rule. We first differentiate [tex]\( \sqrt{u} \)[/tex] with respect to [tex]\( u \)[/tex] and then multiply by the derivative of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}
\][/tex]
2. Differentiate [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
Recall that:
[tex]\[
u = 4 + 3f(x)
\][/tex]
Therefore:
[tex]\[
\frac{du}{dx} = 3 \frac{df(x)}{dx} = 3f'(x)
\][/tex]
Now, we have:
[tex]\[
\frac{d}{dx} \sqrt{4 + 3f(x)} = \frac{1}{2\sqrt{4 + 3f(x)}} \cdot 3f'(x)
\][/tex]
Simplifying:
[tex]\[
h'(x) = \frac{3f'(x)}{2\sqrt{4 + 3f(x)}}
\][/tex]
3. Evaluate [tex]\( h'(x) \)[/tex] at [tex]\( x = 4 \)[/tex]:
We are given [tex]\( f(4) = 4 \)[/tex] and [tex]\( f'(4) = 3 \)[/tex]. Substitute these values into the expression for [tex]\( h'(x) \)[/tex]:
[tex]\[
h'(4) = \frac{3 f'(4)}{2\sqrt{4 + 3 f(4)}}
\][/tex]
Substitute [tex]\( f(4) = 4 \)[/tex] and [tex]\( f'(4) = 3 \)[/tex]:
[tex]\[
h'(4) = \frac{3 \cdot 3}{2\sqrt{4 + 3 \cdot 4}}
\][/tex]
Simplify inside the square root:
[tex]\[
4 + 3 \cdot 4 = 4 + 12 = 16
\][/tex]
Thus:
[tex]\[
h'(4) = \frac{9}{2 \cdot \sqrt{16}} = \frac{9}{2 \cdot 4} = \frac{9}{8}
\][/tex]
Therefore, the derivative [tex]\( h'(4) \)[/tex] is:
[tex]\[
h'(4) = \frac{9}{8}
\][/tex]
