How do you use the chain rule to differentiate #log_(13)cscx#?
Here ,
Using Change of Base Formula:
#color(blue)(log_aX=log_k X/log_ka ,where,k " is the new base"#
So .
Let ,
Using Chain Rule:
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Same as above answer, different method.
Alternatively...
As above.
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To differentiate log base 13 of csc(x) with respect to x using the chain rule, follow these steps:
- Recognize that log base 13 of csc(x) can be expressed as log base 13 of (1/sin(x)).
- Apply the chain rule: differentiate the outer function (log base 13) and then multiply it by the derivative of the inner function (1/sin(x)).
- The derivative of log base 13 of u is (1/u)ln(13), where u = csc(x).
- The derivative of (1/sin(x)) is -csc(x)cot(x).
- Combine the results: (-csc(x)cot(x))(1/csc(x)ln(13)) = -cot(x)ln(13).
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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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