Given the function f(x)=(2x^(2)+3)/(3+5x^(2)), find f^(')(x) in simplified form. Answer: f^(')(x)=

Apply Quotient Rule: To find the derivative of the function f(x) = (2x^2 + 3) / (3 + 5x^2), we will use the quotient rule. The quotient rule states that if we have a function h(x) = u(x) / v(x), then h'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2. Here, u(x) = 2x^2 + 3 and v(x) = 3 + 5x^2.Find u'(x): First, we need to find the derivative of u(x) = 2x^2 + 3. The derivative of 2x^2 is 4x, and the derivative of 3 is 0, so u'(x) = 4x.Find v'(x): Next, we need to find the derivative of v(x) = 3 + 5x^2. The derivative of 5x^2 is 10x, and the derivative of 3 is 0, so v'(x) = 10x.Plug into Quotient Rule: Now we apply the quotient rule. We have u'(x) = 4x and v'(x) = 10x, so we plug these into the quotient rule formula: f'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2 f'(x) = (4x(3 + 5x^2) - (2x^2 + 3)10x) / (3 + 5x^2)^2Simplify Numerator: We simplify the numerator of the derivative: f'(x) = (12x + 20x^3 - 20x^3 - 30x) / (3 + 5x^2)^2 f'(x) = (-18x) / (3 + 5x^2)^2Factor Out -6x: We can simplify the derivative further by factoring out a -6x from the numerator: f'(x) = -6x(3) / (3 + 5x^2)^2 f'(x) = -6x / (3 + 5x^2)

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Given the function
f(x)=(2x^(2)+3)/(3+5x^(2)), find
f^(')(x) in simplified form.
Answer:
f^(')(x)=

Given the function $f(x)=\frac{2 x^{2}+3}{3+5 x^{2}}$ , find $f^{\prime}(x)$ in simplified form. $\newline$ Answer: $f^{\prime}(x)=$

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Algebra 2

Composition of linear and quadratic functions: find an equation

Full solution

Q. Given the function

f(x)=\frac{2 x^{2}+3}{3+5 x^{2}}

, find

f^{\prime}(x)

in simplified form.

\newline

Answer:

f^{\prime}(x)=

Apply Quotient Rule: To find the derivative of the function $f(x) = \frac{2x^2 + 3}{3 + 5x^2}$ , we will use the quotient rule. The quotient rule states that if we have a function $h(x) = \frac{u(x)}{v(x)}$ , then $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ . Here, $u(x) = 2x^2 + 3$ and $v(x) = 3 + 5x^2$ .
Find $u'(x)$ : First, we need to find the derivative of $u(x) = 2x^2 + 3$ . The derivative of $2x^2$ is $4x$ , and the derivative of $3$ is $0$ , so $u'(x) = 4x$ .
Find $v'(x)$ : Next, we need to find the derivative of $v(x) = 3 + 5x^2$ . The derivative of $5x^2$ is $10x$ , and the derivative of $3$ is $0$ , so $v'(x) = 10x$ .
Plug into Quotient Rule: Now we apply the quotient rule. We have $u'(x) = 4x$ and $v'(x) = 10x$ , so we plug these into the quotient rule formula: $\newline$ $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ $\newline$ $f'(x) = \frac{(4x(3 + 5x^2) - (2x^2 + 3)10x)}{(3 + 5x^2)^2}$
Simplify Numerator: We simplify the numerator of the derivative: $\newline$ $f'(x) = \frac{12x + 20x^3 - 20x^3 - 30x}{(3 + 5x^2)^2}$ $\newline$ $f'(x) = \frac{-18x}{(3 + 5x^2)^2}$
Factor Out $-6x$ : We can simplify the derivative further by factoring out a $-6x$ from the numerator: $\newline$ $f'(x) = \frac{-6x(3)}{(3 + 5x^2)^2}$ $\newline$ $f'(x) = \frac{-6x}{(3 + 5x^2)}$