Mathematics, 27.06.2019 13:30 arielgonzalez809

Find dy, dx if f(x) = (x + 8)3x. 3xln(x + 8) the product of the quantity 3 times the natural log of the quantity x plus 8 plus 3 times x divided by the quantity x plus 8, and the quantity x plus 8 raised to the 3x power 3 times the natural log of the quantity x plus 8 plus 3 times x divided by the quantity x plus 8 3x(x + 8)(3x − 1)

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Answer from: NetherisIsTheQueen

$\displaystyle \frac{dy}{dx} = 3(x + 8)^{3x} \bigg[ \ln (x+ 8) + \frac{x}{x + 8} \bigg]$

General Formulas and Concepts:
Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle f(x) = (x + 8)^{3x}$

Step 2: Differentiate

Exponential Differentiation [Derivative Rule - Chain Rule]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \frac{d}{dx} \bigg[ 3x \ln (x + 8) \bigg]$ Derivative Rule [Product Rule]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \bigg[ \frac{d}{dx}(3x) \ln (x + 8) + 3x \frac{d}{dx}[ \ln (x + 8)] \bigg]$ Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \bigg[ 3 \frac{d}{dx}(x) \ln (x + 8) + 3x \frac{d}{dx}[ \ln (x + 8)] \bigg]$ Derivative Rule [Basic Power Rule]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \bigg[ 3 \ln (x + 8) + 3x \frac{d}{dx}[ \ln (x + 8)] \bigg]$ Logarithmic Differentiation [Derivative Rule - Chain Rule]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \bigg[ 3 \ln (x + 8) + \frac{3x}{x + 8} \frac{d}{dx} (x + 8) \bigg]$ Rewrite [Derivative Property - Addition/Subtraction]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \bigg[ 3 \ln (x + 8) + \frac{3x}{x + 8} \Big[ \frac{d}{dx}(x) + \frac{d}{dx}(8) \Big] \bigg]$ Derivative Rule [Basic Power Rule]: $\displaystyle \frac{dy}{dx} = (x + 8)^{3x} \bigg[ 3 \ln (x + 8) + \frac{3x}{x + 8} \bigg]$ Simplify: $\displaystyle \frac{dy}{dx} = 3(x + 8)^{3x} \bigg[ \ln (x+ 8) + \frac{x}{x + 8} \bigg]$

∴ we have found the derivative of the given function.

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Answer from: Quest

linear

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.

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Find dy, dx if f(x) = (x + 8)3x. 3xln(x + 8) the product of the quantity 3 times the natural log of...

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