Question

Find the derivative of $ f(x) = \dfrac{1}{x} $ .

Answer 1

Hint: This is a question from limit and differentiation. To solve the above question we will use the differentiation formula and then putting h as zero we will get the answer.

Complete step-by-step answer:
According to the question, we have the function, $ f(x) = \dfrac{1}{x} $
The derivative of the function $ f(x) $ is denoted as $ f'(x) $ .
According to the limit and continuity formula we know that, the derivative of the function $ f(x) $ is given by,
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $ ……………… (1)
We have given function, $ f(x) = \dfrac{1}{x} $
Hence, Putting (x + h) in place of x we get,
 $ f(x + h) = \dfrac{1}{{x + h}} $
Putting these values in equation 1 we get,
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{x + h}} - \dfrac{1}{x}}}{h} $
Taking L.C.M and simplifying the numerator of the right hand side of the equation we get,
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{x - x - h}}{{(x + h)x}}}}{h} $
Cancelling +x and –x from the numerator of the right hand side of the equation we get,
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{ - h}}{{(x + h)x}}}}{h} $
Again simplifying it we get,
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - h}}{{(x + h)x}} \times \dfrac{1}{h} $
Cancelling h from numerator and denominator of the right hand side of the equation we get,
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 1}}{{(x + h)x}} $
Putting h = 0 in the right hand side of the equation we get,
 $ f'(x) = \dfrac{{ - 1}}{{(x + 0)x}} $
 $ \therefore $ $ f'(x) = \dfrac{{ - 1}}{{{x^2}}} $
 $ \therefore $ The derivative of $ f(x) = \dfrac{1}{x} $ is $ - \dfrac{1}{{{x^2}}} $ .

Note: If you put h = 0 at any step you will not get the answer.
The derivative of a function f at x = c is the limit of the slope of the secant line from x = c to x = c + h as h approaches zero.
From the given problem and its solution, we have got the derivative formula that, if $ f(x) = \dfrac{1}{x} $ then $ f'(x) = \dfrac{{ - 1}}{{{x^2}}} $ .
If we integrate $ f'(x) = \dfrac{{ - 1}}{{{x^2}}} $ , then we will get $ f(x) = \dfrac{1}{x} $ .
You should know all the laws, formulae and the properties of limit and differentiation.
A function is differentiable at a point when there is a defined derivative at that point. The formula of the differentiability is given by, $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $ . Where h is the limit of the function.