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

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

Explanation

Solution

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)=1xf(x) = \dfrac{1}{x}
The derivative of the function f(x)f(x) is denoted as f′(x)f'(x) .
According to the limit and continuity formula we know that, the derivative of the function f(x)f(x) is given by,
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} ……………… (1)
We have given function, f(x)=1xf(x) = \dfrac{1}{x}
Hence, Putting (x + h) in place of x we get,
f(x+h)=1x+hf(x + h) = \dfrac{1}{{x + h}}
Putting these values in equation 1 we get,
f′(x)=lim⁡h→01x+h−1xhf'(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)=lim⁡h→0x−x−h(x+h)xhf'(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)=lim⁡h→0−h(x+h)xhf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{ - h}}{{(x + h)x}}}}{h}
Again simplifying it we get,
f′(x)=lim⁡h→0−h(x+h)x×1hf'(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)=lim⁡h→0−1(x+h)xf'(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)=−1(x+0)xf'(x) = \dfrac{{ - 1}}{{(x + 0)x}}
∴\therefore f′(x)=−1x2f'(x) = \dfrac{{ - 1}}{{{x^2}}}
∴\therefore The derivative of f(x)=1xf(x) = \dfrac{1}{x} is −1x2- \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)=1xf(x) = \dfrac{1}{x} then f′(x)=−1x2f'(x) = \dfrac{{ - 1}}{{{x^2}}} .
If we integrate f′(x)=−1x2f'(x) = \dfrac{{ - 1}}{{{x^2}}} , then we will get f(x)=1xf(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)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} . Where h is the limit of the function.