I know how to find the derivative of a function by itself given a function. I can use multiple rules to determine the derivative. However in a situation where I must find the derivative when the x value is equals to some constant, I am tripped up. For example:
Say I have to find the derivative $\dfrac{dy}{dx}$ for some function:
$y=\dfrac{1}{x}$
I have found that my derivative is $-\dfrac{1}{x^2}$.
How do I now determine $\dfrac{dy}{dx}$ for $y=\dfrac{1}{x}$ when $x = 5$?
This is using the definition of the derivative!
$\endgroup$ 13 Answers
$\begingroup$The easy way is to just plug $x=5$ back into the thing:
$$\frac d{dx}\frac1x\bigg|_{x=5}=-\frac1{x^2}\bigg|_{x=5}=-\frac1{5^2}$$
The harder way is to use the definition:
$$\frac d{dx}\frac1x\bigg|_{x=5}=\lim_{h\to5}\frac{\frac1h-\frac15}{h-5}=\lim_{h\to5}\frac{\frac{5-h}{5h}}{h-5}=-\lim_{h\to5}\frac1{5h}=-\frac1{5^2}$$
$\endgroup$ 4 $\begingroup$The "multiple rules" you use to find the derivative all follow from the definition of the derivative. To use that definition for a particular function $f$ at a particular point, say $x=5$, you must find $$ \lim_{h \rightarrow 5} \frac{f(5+h) -f(5)}{h} . $$
So write that out for the function in your example and do the algebra with fractions to find the limit. Then you can check with the formula you know.
$\endgroup$ $\begingroup$This is quite a strange question indeed.
I guess you know how to calculate $f(x)$ when $x=5$ this is $f(5)$.
Then why are you puzzled by plugging $x=5$ into $f'(x)$, this is just $f'(5)$.
$f(x)=\frac1x$ is a function, $g(x)=-\frac1{x^2}$ is also a function and I'm sure you know how to calculate $g(5)$. Well call it $g$ or call it $f'$ there is no difference in evaluating it for some value of the variable.
Now when $y=1/x$ we are just saying that $y(x)$ is a function of $x$ and $\frac{dy}{dx}$ is nothing more than $y'(x)$.
So the definition of $\frac{dy}{dx}\bigg|_{x=5}$ is as usually written $\lim\limits_{h\to 5}\frac{y(h)-y(5)}{h-5}$.
Is it clearer ?
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