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Math  /  Calculus

QuestionFind the derivative of f(x)=ln⁡1(4x−7)8f(x)=\ln \frac{1}{(4 x-7)^{8}} A. 8(4x−7)8(4 x-7) B. (4x−7)8(4 x-7)^{8} C. −84x−7\frac{-8}{4 x-7} D. −324x−7\frac{-32}{4 x-7} E=1(4x−7)8\left.E=\frac{1}{\left(4 x^{-7}\right.}\right)^{8} Answer!

Studdy Solution

STEP 1

1. We are given the function f(x)=ln⁡1(4x−7)8 f(x) = \ln \frac{1}{(4x-7)^8} .
2. We need to find the derivative f′(x) f'(x) .
3. We will use the chain rule and properties of logarithms to simplify the differentiation process.

STEP 2

1. Simplify the logarithmic expression using properties of logarithms.
2. Differentiate the simplified expression.
3. Simplify the derivative to match one of the given options.

STEP 3

Simplify the logarithmic expression:
Using the property of logarithms: ln⁡1a=−ln⁡a\ln \frac{1}{a} = -\ln a, we have:
f(x)=ln⁡1(4x−7)8=−ln⁡((4x−7)8) f(x) = \ln \frac{1}{(4x-7)^8} = -\ln((4x-7)^8)
Using another property: ln⁡(ab)=bln⁡a\ln(a^b) = b\ln a, we get:
f(x)=−8ln⁡(4x−7) f(x) = -8\ln(4x-7)

STEP 4

Differentiate the simplified expression:
The derivative of −8ln⁡(4x−7)-8\ln(4x-7) is found using the chain rule. The derivative of ln⁡u\ln u is 1u⋅dudx\frac{1}{u} \cdot \frac{du}{dx}.
Let u=4x−7 u = 4x - 7 , then dudx=4\frac{du}{dx} = 4.
Thus, the derivative is:
f′(x)=−8⋅14x−7⋅4 f'(x) = -8 \cdot \frac{1}{4x-7} \cdot 4
f′(x)=−324x−7 f'(x) = \frac{-32}{4x-7}

STEP 5

Simplify the derivative:
The derivative simplifies to:
f′(x)=−324x−7 f'(x) = \frac{-32}{4x-7}
This matches option D.
The correct answer is:
D \boxed{D}

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