How do you differentiate #f(x) = log_3(log_7 (log_5(x + 2)))#?

Answer 1

#f'(x)=1/(ln(3)(x+2)ln(log_5(x+2))ln(x+2))#

The typical method for differentiating a logarithm that doesn't have base #e# requires using the change of base formula to put the expression into terms of the natural logarithm.
For example, the derivative of #log_a(g(x)))# can be found through saying that
#d/dxlog_a(g(x))=d/dx(ln(g(x))/ln(a))=1/ln(a)(d/dxln(g(x)))#
#=1/ln(a)(1/g(x))*g'(x)=1/(g(x)ln(a))*g'(x)#

So, this rule will give us:

#f'(x)=1/(log_7(log_5(x+2))ln(3))*d/dxlog_7(log_5(x+2))#

Using this rule again on the smaller logarithm function (like peeling off its shells):

#f'(x)=1/(log_7(log_5(x+2))ln(3))(1/(log_5(x+2)ln(7)))*d/dxlog_5(x+2)#
Noting that #d/dxlog_5(x+2)=1/((x+2)ln(5))*d/dx(x+2)#, which equals #1/((x+2)ln(5))#, the entire function's derivative equals
#f'(x)=1/(ln(3)ln(5)ln(7)(x+2)log_7(log_5(x+2))log_5(x+2))#
This can be simplified by using the change of base formula on #log_7(log_5(x+2))# and #log_5(x+2))#:
#f'(x)=1/(ln(3)color(red)(cancel(color(black)(ln(5))))color(red)(cancel(color(black)(ln(7))))(x+2)(ln(log_5(x+2))/color(red)(cancel(color(black)(ln(7)))))(ln(x+2)/color(red)(cancel(color(black)(ln(5))))))#
#f'(x)=1/(ln(3)(x+2)ln(log_5(x+2))ln(x+2))#
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Answer 2

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:

  1. Start from the innermost function and differentiate it with respect to ( x ).
  2. 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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Answer from HIX Tutor

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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