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3 years ago

What is log b^b^6x equivalent to (The b is under log and ^b)? How do you know?

Mathematics

1 answer:

Arturiano [62]3 years ago

4 0

Answer: 6x

Work Shown:

For each step, the logs are all base b. This is to save time and hassle of writing tricky notation of having to write the smaller subscript 'b' multiple times. The first rule to use is that log(x^y) = y*log(x) for any base of a logarithm. The second rule is that $\log_b(b) = 1$ meaning that the log base of itself is 1

log(b^(6x)) = 6x*log(b) .... pull down exponent using the first rule above

log(b^(6x)) = 6x*1 .... use the second rule mentioned

log(b^(6x)) = 6x

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Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$

Nothing to do to simplify any further, other than factoring out $x^2$ .

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3 years ago

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Step-by-step explanation:

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M ( x1 + x2/2 , y1 + y2/2)

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M ( -2, 1)

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2 years ago

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Step-by-step explanation:

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katrin2010 [14]

Answer:

A = yes

B = no

C = yes

D = no

E = yes

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