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saul85 [17]
3 years ago
6

Explain how to write 2.14. 10° in standard form by moving the decimal point

Mathematics
1 answer:
pentagon [3]3 years ago
6 0

2.14 * 10^0??

Please be more clear, I'd like to help.

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In Louisiana, the coldest temperature ever recorded is -16°F. That is much warmer than in
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it is 53 degrees warmer than Utah

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Which expression can be used to find 7(-6 1/8)
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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
mote1985 [20]

Answer:

\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}

Step-by-step explanation:

To find the derivative of the function y(x)=\ln \left(\frac{x}{x^2+1}\right) you must:

Step 1. Rewrite the logarithm:

\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }

Step 2. The derivative of a sum is the sum of derivatives:

\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)

Step 3. The derivative of natural logarithm is \left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}

{\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }

Step 4. The function \ln{\left(x^{2} + 1 \right)} is the composition f\left(g\left(x\right)\right) of two functions f\left(u\right)=\ln{\left(u \right)} and u=g\left(x\right)=x^{2} + 1

Step 5.  Apply the chain rule \left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }

-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}

Return to the old variable:

- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}

The derivative of a sum is the sum of derivatives:

- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)

Step 6. Apply the power rule \frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}

\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\

\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}

Thus, \frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}

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Which of the following equations shows how substitution can be used to solve the following
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Step-by-step explanation:

the  y=   equation shows how... b/c we can just plug y into the 2nd equation and eliminate y and have one variable, x , to solve for  :)

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