How do you differentiate #f(x) = log_3(log_7 (log_5(x + 2)))#?
So, this rule will give us:
Using this rule again on the smaller logarithm function (like peeling off its shells):
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To differentiate the function ( f(x) = \log_3(\log_7(\log_5(x + 2))) ), you can use the chain rule of differentiation. Here's the step-by-step process:
- Start from the innermost function and differentiate it with respect to ( x ).
- Then proceed to the outer functions, working your way outwards.
Let's denote: [ u = \log_5(x + 2) ] [ v = \log_7(u) ] [ f(x) = \log_3(v) ]
Now, differentiate each step:
[ \frac{du}{dx} = \frac{1}{(x + 2)\ln(5)} ] [ \frac{dv}{du} = \frac{1}{u\ln(7)} ] [ \frac{df}{dv} = \frac{1}{v\ln(3)} ]
Now, apply the chain rule:
[ \frac{df}{dx} = \frac{df}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} ]
Substitute the derivatives we calculated:
[ \frac{df}{dx} = \frac{1}{v\ln(3)} \cdot \frac{1}{u\ln(7)} \cdot \frac{1}{(x + 2)\ln(5)} ]
[ \frac{df}{dx} = \frac{1}{(x + 2)u\ln(3)\ln(5)\ln(7)} ]
Finally, substitute back the expressions for ( u ) and ( v ):
[ \frac{df}{dx} = \frac{1}{(x + 2)\log_5(x + 2)\ln(3)\ln(5)\ln(7)} ]
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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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