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(1 point) Suppose that the functions [tex]$f$[/tex] and [tex]$g$[/tex] and their derivatives have the following values at [tex]$x=0$[/tex] and [tex]$x=1$[/tex].

[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) \\
\hline
0 & 1 & 1 & -2.4 & 6.6 \\
\hline
1 & -1.8 & 6.9 & -1.3 & -2.3 \\
\hline
\end{array}
\][/tex]

Find each of the following derivatives.

[tex]\[
\begin{array}{l}
\left.\frac{d}{d x}(2 f(x)-g(x))\right|_{x=1}=-0.3 \\
\left.\frac{d}{d x}\left(f(x)(g(x))^2\right)\right|_{x=0}=10.8 \\
\left.\frac{d}{d x} \frac{f(x)}{g(x)+2}\right|_{x=1}= \\
\left.\frac{d}{d x} f(g(x))\right|_{x=0}=-8.58 \\
\left.\frac{d}{d x} g(f(x))\right|_{x=0}=5.52 \\
\left.\frac{d}{d x}\left(x^9+f(x)\right)^{-9}\right|_{x=1}=-645.4 \\
\left.\frac{d}{d x} f(x+g(x))\right|_{x=0}=-9.88
\end{array}
\][/tex]


Sagot :

### Calculating [tex]\(\left. \frac{d}{dx} \frac{f(x)}{g(x)+2} \right|_{x=1} \)[/tex]

To find the derivative of [tex]\(\frac{f(x)}{g(x)+2}\)[/tex] at [tex]\(x=1\)[/tex], we'll first express the function and then apply the quotient rule.

The quotient rule states that for two differentiable functions [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex],

[tex]\[ \left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \][/tex]

In our case:
- [tex]\( u(x) = f(x) \)[/tex], so [tex]\( u'(x) = f'(x) \)[/tex]
- [tex]\( v(x) = g(x) + 2 \)[/tex], so [tex]\( v'(x) = g'(x) \)[/tex]

### Step-by-Step Calculation:

1. Evaluate [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex] at [tex]\(x=1\)[/tex]:
[tex]\[ u(1) = f(1) = -1.8 \][/tex]
[tex]\[ v(1) = g(1) + 2 = 6.9 + 2 = 8.9 \][/tex]

2. Evaluate [tex]\(u'(x)\)[/tex] and [tex]\(v'(x)\)[/tex] at [tex]\(x=1\)[/tex]:
[tex]\[ u'(1) = f'(1) = -1.3 \][/tex]
[tex]\[ v'(1) = g'(1) = -2.3 \][/tex]

3. Apply the quotient rule:
[tex]\[ \left( \frac{f(x)}{g(x) + 2} \right)' \Bigg|_{x=1} = \frac{f'(x)(g(x)+2) - f(x)g'(x)}{(g(x)+2)^2} \Bigg|_{x=1} \][/tex]

4. Substituting the values:
[tex]\[ \left( \frac{f(x)}{g(x)+2} \right)' \Bigg|_{x=1} = \frac{-1.3 \cdot 8.9 - (-1.8) \cdot (-2.3)}{8.9^2} \][/tex]

5. Simplify the numerator:
[tex]\[ -1.3 \cdot 8.9 = -11.57 \][/tex]
[tex]\[ -1.8 \cdot -2.3 = 4.14 \][/tex]
[tex]\[ -11.57 - 4.14 = -15.71 \][/tex]

6. Simplify the denominator:
[tex]\[ 8.9^2 = 79.21 \][/tex]

7. Combine to find the derivative:
[tex]\[ \left. \frac{d}{dx} \frac{f(x)}{g(x)+2} \right|_{x=1} = \frac{-15.71}{79.21} \approx -0.1983335437444767 \][/tex]

So the derivative [tex]\(\left. \frac{d}{dx} \frac{f(x)}{g(x)+2} \right|_{x=1}\)[/tex] is approximately [tex]\(-0.1983\)[/tex].