(d)/(dx)[sqrt(8x^(2)+2x-3)]=? Choose 1 answer: (A) (1)/(2sqrt(16 x+2)) (B) (8x+1)/(sqrtx) (C) sqrt(16 x+2) (D) (8x+1)/(sqrt(8x^(2)+2x-3))

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(d)/(dx)[sqrt(8x^(2)+2x-3)]=? Choose 1 answer: (A)  (1)/(2sqrt(16 x+2)) (B)  (8x+1)/(sqrtx) (C)  sqrt(16 x+2) (D)  (8x+1)/(sqrt(8x^(2)+2x-3))

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Chain Rule Explanation: To find the derivative of the function $ f(x) = \sqrt{8x^2 + 2x - 3} $, we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.Define Inner and Outer Functions: Let's denote the inner function as $ u(x) = 8x^2 + 2x - 3 $. The outer function is the square root function, which we can write as $ v(u) = \sqrt{u} $. We need to find the derivatives $ v'(u) $ and $ u'(x) $.Find Derivative of Outer Function: The derivative of the outer function $ v(u) = \sqrt{u} $ with respect to $ u $ is $ v'(u) = \frac{1}{2\sqrt{u}} $.Find Derivative of Inner Function: The derivative of the inner function $ u(x) = 8x^2 + 2x - 3 $ with respect to $ x $ is $ u'(x) = 16x + 2 $.Apply Chain Rule: Now we apply the chain rule: $ \frac{d}{dx} f(x) = v'(u) \cdot u'(x) $. Substituting the derivatives we found, we get $ \frac{d}{dx} f(x) = \frac{1}{2\sqrt{u}} \cdot (16x + 2) $.Substitute Derivatives: Substitute back $ u $ with $ 8x^2 + 2x - 3 $ to express the derivative in terms of $ x $: $ \frac{d}{dx} f(x) = \frac{1}{2\sqrt{8x^2 + 2x - 3}} \cdot (16x + 2) $.Simplify Expression: Simplify the expression: $ \frac{d}{dx} f(x) = \frac{16x + 2}{2\sqrt{8x^2 + 2x - 3}} $.Final Derivative: Further simplify the expression by dividing the numerator and the denominator by 2: $ \frac{d}{dx} f(x) = \frac{8x + 1}{\sqrt{8x^2 + 2x - 3}} $.

$\frac{d}{d x}\left[\sqrt{8 x^{2}+2 x-3}\right]=?$ $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2 \sqrt{16 x+2}}$ $\newline$ (B) $\frac{8 x+1}{\sqrt{x}}$ $\newline$ (C) $\sqrt{16 x+2}$ $\newline$ (D) $\frac{8 x+1}{\sqrt{8 x^{2}+2 x-3}}$

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