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10 months ago

How do you find the derivative of 1/logx?

Mathematics

1 answer:

valentina_108 [34]10 months ago

5 0

The derivative of 1/logx is With the chain rule.

1log(x)=log(x)−1 is ,= -1xlog(x)2 .

The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".

The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

$With the chain rule. \\ 1log(x)=log(x)−1 \\ So you follow the chain rule to the electricity after which to the log: \\ ddxlog(x)−1=−1⋅log(x) \\ \\−2⋅ddxlog(x) \\ =−1log(x)21x \\ =−1xlog(x)2 .$

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Answer:

Step-by-step explanation:

Domain x^2 - 9 {Solution: - infinity < x < infinity}

Interval notation (- infinity, infinity)

Range of x^2 - 9 (Solution: f(x) is greater than or equal to - 9)

Interval notation (-9, infinity)

Axis interception points of x^2 - 9:

X- intercepts (3, 0) (-3, 0)

Y-intercepts (0, -9)

Vertex of x^2 - 9: Minimum (0, -9)

Solve for f:

f (x) = x^2 - 9

Step 1: Divide both sides by x.

fx / x = x^2 - 9 / x

f = x^2 - 9 / x

Answer:

f = x^2 - 9 / x

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2 years ago

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attashe74 [19]

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2 years ago

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Answer:

A = 2,384.1708

Step-by-step explanation:

A = P(1 + rt)

P = 942.36

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t = 12 months

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3 years ago