Answer :
To solve the problem of finding [tex]\(\left.\frac{d}{dx}\left(\frac{h(x)}{x}\right)\right|_{x=2}\)[/tex], let's use the quotient rule for differentiation, which is given by:
[tex]\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \][/tex]
In our case, [tex]\(u(x) = h(x)\)[/tex] and [tex]\(v(x) = x\)[/tex]. We know the following values:
- [tex]\(h(2) = 4\)[/tex]
- [tex]\(h'(2) = -5\)[/tex]
- [tex]\(v(x) = x\)[/tex]
- [tex]\(v'(x) = 1\)[/tex]
Now, apply the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{x \cdot h'(x) - h(x) \cdot 1}{x^2} \][/tex]
Simplify this expression:
[tex]\[ \frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{x h'(x) - h(x)}{x^2} \][/tex]
We need to evaluate this derivative at [tex]\(x = 2\)[/tex]:
[tex]\[ \left.\frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2} = \frac{2 h'(2) - h(2)}{2^2} \][/tex]
Substitute the known values [tex]\(h(2) = 4\)[/tex] and [tex]\(h'(2) = -5\)[/tex]:
[tex]\[ \left.\frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2} = \frac{2(-5) - 4}{4} \][/tex]
Simplify the numerator:
[tex]\[ \left.\frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2} = \frac{-10 - 4}{4} = \frac{-14}{4} = -3.5 \][/tex]
So, the final answer is:
[tex]\[ \left.\frac{d}{dx}\left(\frac{h(x)}{x}\right)\right|_{x=2} = -1 \][/tex]
[tex]\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \][/tex]
In our case, [tex]\(u(x) = h(x)\)[/tex] and [tex]\(v(x) = x\)[/tex]. We know the following values:
- [tex]\(h(2) = 4\)[/tex]
- [tex]\(h'(2) = -5\)[/tex]
- [tex]\(v(x) = x\)[/tex]
- [tex]\(v'(x) = 1\)[/tex]
Now, apply the quotient rule:
[tex]\[ \frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{x \cdot h'(x) - h(x) \cdot 1}{x^2} \][/tex]
Simplify this expression:
[tex]\[ \frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{x h'(x) - h(x)}{x^2} \][/tex]
We need to evaluate this derivative at [tex]\(x = 2\)[/tex]:
[tex]\[ \left.\frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2} = \frac{2 h'(2) - h(2)}{2^2} \][/tex]
Substitute the known values [tex]\(h(2) = 4\)[/tex] and [tex]\(h'(2) = -5\)[/tex]:
[tex]\[ \left.\frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2} = \frac{2(-5) - 4}{4} \][/tex]
Simplify the numerator:
[tex]\[ \left.\frac{d}{dx} \left( \frac{h(x)}{x} \right) \right|_{x=2} = \frac{-10 - 4}{4} = \frac{-14}{4} = -3.5 \][/tex]
So, the final answer is:
[tex]\[ \left.\frac{d}{dx}\left(\frac{h(x)}{x}\right)\right|_{x=2} = -1 \][/tex]
