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Given the function
y=(2x^(2))/(6-5x^(2)), find
(dy)/(dx) in simplified form.
Answer:
(dy)/(dx)=

Given the function $y=\frac{2 x^{2}}{6-5 x^{2}}$ , find $\frac{d y}{d x}$ in simplified form. $\newline$ Answer: $\frac{d y}{d x}=$

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Write a formula for an arithmetic sequence

Full solution

Q. Given the function

y=\frac{2 x^{2}}{6-5 x^{2}}

, find

\frac{d y}{d x}

in simplified form.

\newline

Answer:

\frac{d y}{d x}=

Identify Function: Identify the function that needs differentiation. $\newline$ We are given the function $y=\frac{2x^{2}}{6-5x^{2}}$ . We need to find its derivative with respect to $x$ , which is denoted as $\frac{dy}{dx}$ .
Apply Quotient Rule: Apply the quotient rule for differentiation. $\newline$ The quotient rule states that if we have a function in the form of $\frac{u(x)}{v(x)}$ , then its derivative is given by $\frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}$ . Here, $u(x) = 2x^2$ and $v(x) = 6 - 5x^2$ .
Differentiate $u(x)$ : Differentiate $u(x)$ with respect to $x$ . The derivative of $u(x) = 2x^2$ with respect to $x$ is $u'(x) = 2 \times 2x = 4x$ .
Differentiate $v(x)$ : Differentiate $v(x)$ with respect to $x$ . The derivative of $v(x) = 6 - 5x^2$ with respect to $x$ is $v'(x) = 0 - 5 \times 2x = -10x$ .
Apply Quotient Rule: Apply the quotient rule using the derivatives from steps $3$ and $4$ . $\newline$ $\frac{dy}{dx} = \frac{(6 - 5x^2) \cdot (4x) - (2x^2) \cdot (-10x)}{(6 - 5x^2)^2}$ .
Simplify Expression: Simplify the expression. $\newline$ $(\frac{dy}{dx}) = \frac{24x - 20x^3 + 20x^3}{36 - 60x^2 + 25x^4}$ . $\newline$ Notice that $-20x^3$ and $+20x^3$ cancel each other out. $\newline$ $(\frac{dy}{dx}) = \frac{24x}{36 - 60x^2 + 25x^4}$ .
Further Simplify: Further simplify the expression if possible. $\newline$ In this case, the expression cannot be simplified further.

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