Let g be a function such that g(4)=16 and g^(')(4)=12. Let h be the function h(x)=sqrtx. Let H be a function defined as H(x)=(g(x))/(h(x)). H^(')(4)=

Write H(x) function: First, let's write down the function H(x) using the given functions g(x) and h(x): H(x) = g(x) / h(x). Find derivative H'(x): Now, we need to find the derivative of H(x), which is H'(x). We'll use the quotient rule for derivatives, which is (f/g)' = (f'g - fg') / g^2, where f is the numerator and g is the denominator. Derivatives of g(x) and h(x): Let's find the derivatives of g(x) and h(x). We know g'(4) = 12. For h(x) = sqrt(x), the derivative h'(x) is 1/(2sqrt(x)). So h'(4) = 1/(2sqrt(4)) = 1/4. Apply quotient rule: Now we apply the quotient rule. H'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2. Plug in x=4: Let's plug in the values for x=4. H'(4) = (g'(4) * h(4) - g(4) * h'(4)) / (h(4))^2. Substitute known values: Substitute the known values: H'(4) = (12 * sqrt(4) - 16 * 1/4) / (sqrt(4))^2. Calculate values: Calculate the values: H'(4) = (12 * 2 - 16 * 1/4) / 4. Simplify expression: Simplify the expression: H'(4) = (24 - 4) / 4. Finish calculation: Finish the calculation: H'(4) = 20 / 4. Final result: So, H'(4) = 5.

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Let
g be a function such that
g(4)=16 and
g^(')(4)=12.
Let
h be the function
h(x)=sqrtx.

Let
H be a function defined as
H(x)=(g(x))/(h(x)).

H^(')(4)=

- Let $g$ be a function such that $g(4)=16$ and $g^{\prime}(4)=12$ . $\newline$ - Let $h$ be the function $h(x)=\sqrt{x}$ . $\newline$ Let $H$ be a function defined as $H(x)=\frac{g(x)}{h(x)}$ . $\newline$ $H^{\prime}(4)=$

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Math Problems

Calculus

Find derivatives using logarithmic differentiation

Full solution

Q. - Let

g

be a function such that

g(4)=16

and

g^{\prime}(4)=12

\newline

- Let

h

be the function

h(x)=\sqrt{x}

\newline

Let

H

be a function defined as

H(x)=\frac{g(x)}{h(x)}

\newline

H^{\prime}(4)=

Write $H(x)$ function: First, let's write down the function $H(x)$ using the given functions $g(x)$ and $h(x)$ : $H(x) = \frac{g(x)}{h(x)}$ .
Find derivative $H'(x)$ : Now, we need to find the derivative of $H(x)$ , which is $H'(x)$ . We'll use the quotient rule for derivatives, which is $(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ , where $f$ is the numerator and $g$ is the denominator.
Derivatives of $g(x)$ and $h(x)$ : Let's find the derivatives of $g(x)$ and $h(x)$ . We know $g'(4) = 12$ . For $h(x) = \sqrt{x}$ , the derivative $h'(x)$ is $1/(2\sqrt{x})$ . So $h'(4) = 1/(2\sqrt{4}) = 1/4$ .
Apply quotient rule: Now we apply the quotient rule. $H'(x) = \frac{g'(x) \cdot h(x) - g(x) \cdot h'(x)}{(h(x))^2}$ .
Plug in $x=4$ : Let's plug in the values for $x=4$ . $H'(4) = \frac{g'(4) \cdot h(4) - g(4) \cdot h'(4)}{(h(4))^2}$ .
Substitute known values: Substitute the known values: $H'(4) = (12 \times \sqrt{4} - 16 \times \frac{1}{4}) / (\sqrt{4})^2$ .
Calculate values: Calculate the values: $H'(4) = \frac{(12 \times 2 - 16 \times \frac{1}{4})}{4}.$
Simplify expression: Simplify the expression: $H'(4) = \frac{24 - 4}{4}.$
Finish calculation: Finish the calculation: $H'(4) = \frac{20}{4}.$
Final result: So, $H'(4) = 5$ .