How do you integrate #f(x)=log_6(2x)/lnx# using the quotient rule?
The quotient rule is only for derivatives, it can't be applied to integrals.
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To integrate ( f(x) = \frac{\log_6(2x)}{\ln(x)} ) using the quotient rule, we first express it as the difference of two functions, ( u(x) ) and ( v(x) ), where ( u(x) = \log_6(2x) ) and ( v(x) = \ln(x) ). Then, we apply the quotient rule, which states that the derivative of ( \frac{u(x)}{v(x)} ) is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ).
First, we find the derivatives of ( u(x) ) and ( v(x) ). The derivative of ( \log_6(2x) ) with respect to ( x ) is ( \frac{1}{\ln(6)} \cdot \frac{1}{2x} ) by the chain rule. The derivative of ( \ln(x) ) is ( \frac{1}{x} ).
Now, we apply the quotient rule:
[ f'(x) = \frac{\left(\frac{1}{\ln(6)} \cdot \frac{1}{2x}\right) \cdot \ln(x) - \log_6(2x) \cdot \frac{1}{x}}{(\ln(x))^2} ]
[ = \frac{\frac{1}{2x \cdot \ln(6)} \cdot \ln(x) - \frac{\log_6(2x)}{x}}{(\ln(x))^2} ]
[ = \frac{\ln(x) - \log_6(2x) \cdot \ln(x)}{2x \cdot \ln(6) \cdot (\ln(x))^2} - \frac{\log_6(2x)}{x \cdot (\ln(x))^2} ]
This is the derivative of ( f(x) ). To find the integral of ( f(x) ), we integrate ( f'(x) ) with respect to ( x ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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