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aev [14]
3 years ago
12

I don't understand this question can someone help me with this please, please explain!! and thank you for the help!! If u dont k

now try your best :D

Mathematics
1 answer:
hammer [34]3 years ago
4 0

Answer:

The answer is B) 2/3x=5 and E) 6x-3+8=11.

Step-by-step explanation:

Basically, all we have to do is see which of the answer options contains an equal sign. So lets go from there:

Does A have an equal sign? We can see the plus sign, a coefficient, and a variable, but we don't have an equal sign, so A is not an equation.

What about B? Well, we can see an equal sign that is showing that two values of an expression equal each other, so yes, B is an equation.

What about C? Well, we can see the minus sign, a variable, and a coefficient, but there is no equal sign. Therefore C is not an equation.

What about D?  Well, D has parenthesis, variables, coefficients, and a plus sign, but no equal sign. So no, D is not an equation.

Finally, is E and equation? Well, we can see that there are two values that are equal to each other as is shown by an equals sign. So yes, E is an equation.

*Had to replace so I could submit the answer

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Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^
serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'

=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'

=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'

=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}

=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}

Note that writing \log(\ln x)^3=3\log(\ln x) is one way to avoid using the power rule.

(2)

\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'

=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'

=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}

=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}

(3)

\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'

=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')

=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})

=3x^2e^{3x}(1+x)\cos(x^3e^{3x})

(4)

\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'

=3\sin^2(xe^x)\cos(xe^x)(xe^x)'

=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')

=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)

=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)

(5) Use implicit differentiation here.

(\ln(xy))'=(e^{2y})'

\dfrac{(xy)'}{xy}=2e^{2y}y'

\dfrac{x'y+xy'}{xy}=2e^{2y}y'

y+xy'=2xye^{2y}y'

y=(2xye^{2y}-x)y'

y'=\dfrac y{2xye^{2y}-x}

8 0
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