For function like $f(x) = \log x,$ the derivative is finite for $x< 0.$ How can slope exist when there is no curve for x< 0 for our function?

For a function f(x) = logx , the derivative is finite for x< 0 . How can slope exist when there is no curve for x< 0 for our function???

Solution 1:

Perhaps you want to consider $f(x):=\ln|x|$. This $f$ is defined for all non-zero reals, is also continuous and differentiable there, with derivative $f’(x)=\frac1x$.

Solution 2:

We are used to say that $\dfrac{d(\log)}{dx}(x) = \dfrac{1}{x}$. But we have to be careful with this, since it is only valid for $x>0$.

By definition of derivative of a function $f$, $$f'(x) = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$ Applying this to $\log$ for $x<0$, $$(\log)'(x) = \lim_{h\to 0} \dfrac{\log(x+h)-\log(x)}{h} $$ and, of course, $\log(x+h)$ and $\log(x)$ don't exist.

Solution 3:

I am also learning mathematics, so my answer might not be correct; I hope someone can correct it if it is wrong. I think the answer is very simple. Functions such as f(x)=log(x) take a value x, in the domain and map it to a value, f(x), in the codomain. The derivative is a function that is based off of the domain for x and map it to f'(x) in a possibly different codomain. The domain of log x is (0, infinity). The derivative is 1/x sure, but it is not f'(x) unless the restriction of x is between (0, infinity) is applied. Otherwise it can no longer be considered the derivative for log x, but of a different function. The version of f'(x) with the entire (-infinity, 0) union (0,infinity) domain is the derivative of a piecewise function that should be quite obvious (include a reflection of log x via piecewise). Hope this makes sense and helps.