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What is the derivative of #1/logx#?

Answer 1

Evelyn Agard

It is #(-1)/(x(logx)^2(ln10)# (Assuming that #logx = log_10x#).

The function #f(x) = 1/logx#

is of the form: #f(x) = 1/u = u^-1#

So we find its derivative using the power rule and the chain rule:

#f'(x) = -1u^-2 * (du)/dx = (-1)/u^2 * (du)/dx#

In this case, #u = logx#, which I take to mean #log_10x#.

The derivative of #log_bx = 1/(xlnb)#, so we have:

#(du)/dx d/dx(logx) = d/dx(log_10 x)= 1/(xln10)#.

Putting this together, we get:

#f'(x) = (-1)/(logx)^2 * 1/(xln10) = (-1)/(x(logx)^2(ln10)#

Note If we are using #logx# for the natural logarithm (as is sometimes done), then, that would be #log_e x# and natural log of #e# is #1#, so we get

#f'(x) = (-1)/(logx)^2 * 1/x = (-1)/(x(logx)^2)#

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Answer 2

Aurora Alvarado

The derivative of ( \frac{1}{\log(x)} ) with respect to ( x ) is ( -\frac{1}{x \cdot \log(x)^2} ).

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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.