Just need someone to clarify things with me!
As I was solving this equation, I got:
x^-2
-------
x^10
I moved x^-2 down with x^10
So in x^-2's place, I placed a positive 1.
1
----------------------
x^10 times x^-2
The radical rule here is to pretty much subtract ahead. However, this extremely smart person said something like if I move a negative number to the denominator, it becomes positive. However, in this case, x^-2 stays the same.
Anyone explain here??
Answers
Answer:
[tex]\frac{1}{x^{12} }[/tex]
Step-by-step explanation:
Given [tex](\frac{x}{x^{-5}})^{-2}[/tex]
The Negative Exponent Rule states that, [tex]a^{-n} = \frac{1}{a^{n}}[/tex]
Basically, since you're raising the fraction inside the parenthesis to a negative exponent, the whole fraction becomes a denominator of 1: [tex]\frac{1}{(\frac{x}{x^{-5}})^{2}}[/tex]
Then, you'll have to work on the denominator:
[tex]\frac{1}{(\frac{x}{x^{-5}})^{2}} = \frac{1}{\frac{x^{2} }{(x^{-5})^{2}}}[/tex]
Next, you'll have to work on the divisor of x² (in the denominator), [tex](x^{-5})^{2}[/tex] by using the Power-to-Power Rule of Exponents: [tex](a^{m})^{n} = a^{mn}[/tex], which results in:
[tex](x^{-5})^{2} = x^{-10}[/tex]
Since it is a negative exponent, you'll have to apply the Negative Exponent Rule once again to the denominator.
[tex]\frac{1}{(\frac{x}{x^{-5}})^{2}} = \frac{1}{\frac{x^{2} }{(x^{-5})^{2}}} = \frac{1}{\frac{x^{2} }{x^{-10}}} = \frac{1}{\frac{x^{2} }{\frac{1}{x^{10}}}}[/tex]
At this point, you could apply the Fraction rule onto the denominator: [tex]\frac{a}{\frac{b}{c}} = \frac{a*c}{b}[/tex]
So the denominator now becomes:
[tex]\frac{1}{\frac{x^{2} }{\frac{1}{x^{10}}}} = \frac{1}{x^{2}*x^{10} }[/tex]
Finally, you could apply the Product Rule of Exponents, [tex]a^{m}a^{n} = a^{m + n }[/tex] onto the denominator:
[tex]\frac{1}{x^{2}*x^{10} } = \frac{1}{x^{2+10}} = \frac{1}{x^{12} }[/tex]
Therefore, the correct answer is: [tex]\frac{1}{x^{12} }[/tex]
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lý thuyết mạch ... giúp em với
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...........................
how to solve this derivative.
f(x)={ln(3x)}^2x
with steps...
Answers
Rewrite f(x) as a nested exponential-logarithm expression :
[tex]\left(\ln(3x)\right)^{2x} = \exp\left(\ln\left(\ln(3x)\right)^{2x}\right)[/tex]
(where [tex]\exp(x) = e^x[/tex])
One of the properties of logarithms lets us drop the exponent as a coefficient:
[tex]\exp\left(\ln\left(\ln(3x)\right)^{2x}\right) = \exp\left(2x\ln\left(\ln(3x)\right)\right)[/tex]
Now, by the chain rule, we have
[tex]f(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \implies \\\\ f'(x) = \left(\exp\left(2x\ln\left(\ln(3x)\right)\right)\right)' \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2x\ln\left(\ln(3x)\right)\right)'[/tex]
By the product rule,
[tex]f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left( (2x)' \ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left( 2\ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)' \right)[/tex]
By the chain rule again,
[tex]f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \dfrac1{\ln(3x)} \cdot \left(\ln(3x)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2x}{\ln(3x)} \cdot \dfrac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{3\ln(3x)} \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)[/tex]
Then simplify this to
[tex]f'(x) = \boxed{\left(\ln(3x)\right)^{2x} \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)}[/tex]
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Answers
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hope this helps have a great day
mark brainliest:)
[tex]\rm\purple{\overbrace{\underbrace{\tt\color{black}{\:\:\:\:\:\:\:\:Answer:\:\:\:\:\:\:\:\:\:}}}}[/tex]
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