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3 years ago

Which of the following illustration the product rule for logarithms equations?

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

4 0

The answer is the last one...the multiply will be (+)

Valentin [98]3 years ago

3 0

I have removed my answer.

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Write as a Improper fraction

Likurg_2 [28]

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8 0

3 years ago

Read 2 more answers

CAN SOMEONE HELP???????????

guapka [62]

Answer:

g(-1) = -1

g(-0.25) = 0

g(2) = 2

Step-by-step explanation:

6 0

3 years ago

What is an insurance premium?

rjkz [21]

Answer: An insurance premium is the amount of money an individual or business pays for an insurance policy. Insurance premiums are paid for policies that cover healthcare, auto, home, life, and others.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

The graph shows that is translated horizontally and vertically to create the function .

yarga [219]

According to the function transformations, the value of h is -2

<h3>How to determine the value of h?</h3>

The complete question is in the attachment

The functions are given as:

$f(x) = (2.5)^x$

$g(x) = (2.5)^{x-h$

From the question, we understand that the function f(x) is translated to the left to get g(x)

From the attached graph, we can see that the function h(x) is 2 units to the left of f(x).

This transformation is represented by:

(x, y) => (x + 2, y)

So, we have:

x - h = x + 2

Evaluate the like terms

h = -2

Hence, the value of h is -2

Read more about function transformations at:

brainly.com/question/3381225

#SPJ1

3 0

2 years ago