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Use the following information:

[tex]\[
\begin{array}{ll}
f(9)=2 & g(9)=3 \\
f^{\prime}(9)=1 & g^{\prime}(9)=4
\end{array}
\][/tex]

Find the following derivatives at [tex]\( x=9 \)[/tex]:

1. [tex]\(\frac{d}{dx}\left(4[f(x)]^2 - 4f(x) + 2\right)\)[/tex]
2. [tex]\(\frac{d}{dx}\left(\frac{1}{3 - 7g(x)}\right)\)[/tex]

[tex]\(\square\)[/tex]


Sagot :

Certainly! Let's solve the given problem step-by-step using the provided information.

### 1. Find [tex]\(\frac{d}{dx}\left(4[f(x)]^2 - 4f(x) + 2\right) \)[/tex] at [tex]\( x = 9 \)[/tex]:

Given:
- [tex]\( f(9) = 2 \)[/tex]
- [tex]\( f'(9) = 1 \)[/tex]

We need to find the derivative of the expression [tex]\( 4[f(x)]^2 - 4f(x) + 2 \)[/tex].

#### Step-by-Step Derivation:

1. Note the given expression: [tex]\( 4[f(x)]^2 - 4f(x) + 2 \)[/tex].

2. Differentiate term by term:
- For the first term [tex]\( 4[f(x)]^2 \)[/tex]:
- Use the chain rule: [tex]\(\frac{d}{dx}[4[f(x)]^2] = 4 \cdot 2f(x) \cdot f'(x) = 8f(x)f'(x) \)[/tex].
- For the second term [tex]\( -4f(x) \)[/tex]:
- Use the chain rule: [tex]\(\frac{d}{dx}[-4f(x)] = -4f'(x)\)[/tex].
- The third term [tex]\( +2 \)[/tex] is a constant, so its derivative is 0.

3. Combine the results:
[tex]\[ \frac{d}{dx}\left(4[f(x)]^2 - 4f(x) + 2\right) = 8f(x)f'(x) - 4f'(x) \][/tex]

4. Substitute [tex]\( x = 9 \)[/tex]:
[tex]\[ \text{At } x = 9, \text{ we have } f(9) = 2 \text{ and } f'(9) = 1. \][/tex]

[tex]\[ \text{Substitute these values in: } 8f(9)f'(9) - 4f'(9) = 8 \cdot 2 \cdot 1 - 4 \cdot 1 = 16 - 4 = 12. \][/tex]

[tex]\[ \boxed{12} \][/tex]

### 2. Find [tex]\(\frac{d}{dx} \left( \frac{1}{3 - 7g(x)} \right) \)[/tex] at [tex]\( x = 9 \)[/tex]:

Given:
- [tex]\( g(9) = 3 \)[/tex]
- [tex]\( g'(9) = 4 \)[/tex]

We need to find the derivative of the expression [tex]\( \frac{1}{3 - 7g(x)} \)[/tex].

#### Step-by-Step Derivation:

1. Note the given expression: [tex]\( \frac{1}{3 - 7g(x)} \)[/tex].

2. Differentiate using the chain rule:
- Use the chain rule: If [tex]\( h(x) = 3 - 7g(x) \)[/tex], then [tex]\( \frac{d}{dx} \left( \frac{1}{h(x)} \right) = -\frac{1}{(h(x))^2} \cdot h'(x) \)[/tex].
- Let [tex]\( h(x) = 3 - 7g(x) \)[/tex] so then [tex]\( h'(x) = -7g'(x) \)[/tex].

3. Combine the results:
[tex]\[ \frac{d}{dx} \left( \frac{1}{3 - 7g(x)} \right) = -\frac{1}{(3 - 7g(x))^2} \cdot (-7g'(x)) = \frac{7g'(x)}{(3 - 7g(x))^2} \][/tex]

4. Substitute [tex]\( x = 9 \)[/tex]:
[tex]\[ \text{At } x = 9, \text{ we have } g(9) = 3 \text{ and } g'(9) = 4. \][/tex]

[tex]\[ \text{Substitute these values in: } \frac{7g'(9)}{(3 - 7g(9))^2} = \frac{7 \cdot 4}{(3 - 7 \cdot 3)^2} = \frac{28}{(3 - 21)^2} = \frac{28}{(-18)^2} = \frac{28}{324} = \frac{7}{81} \][/tex]

[tex]\[ \boxed{0.08641975308641975} \][/tex]

So the results are:
1. [tex]\(\frac{d}{dx}\left(4[f(x)]^2 - 4f(x) + 2\right) \)[/tex] at [tex]\( x = 9 \)[/tex] is [tex]\( \boxed{12} \)[/tex].
2. [tex]\(\frac{d}{dx}\left(\frac{1}{3-7 g(x)}\right) \)[/tex] at [tex]\( x = 9 \)[/tex] is [tex]\( \boxed{0.08641975308641975} \)[/tex].