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Question

Question: How do I find the derivative of \(\dfrac{1}{x}\) using the difference quotient?...

How do I find the derivative of 1x\dfrac{1}{x} using the difference quotient?

Explanation

Solution

We start solving the problem by representing the given equation with a function. We then apply the difference quotient method as the derivative of a function f(x)f\left( x \right) as limh0f(x+h)f(x)h\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} to proceed through the problem. We then make the necessary calculations and make use of the fact that limxaf(x)=f(a)\displaystyle \lim_{x\to a}f\left( x \right)=f\left( a \right) to proceed further through the problem. We then make the necessary calculations to get the required result of derivative of the given function.

Complete step-by-step answer:
According to the problem, we are asked to find the derivative of 1x\dfrac{1}{x} using the difference quotient.
Let us assume f(x)=1xf\left( x \right)=\dfrac{1}{x} ---(1).
From difference quotient method, we know that the derivative of a function f(x)f\left( x \right) is defined as limh0f(x+h)f(x)h\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}. Let us use this result in equation (1).
So, we have f(x)=limh01x+h1xh{{f}^{'}}\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}.
f(x)=limh0x(x+h)x(x+h)h\Rightarrow {{f}^{'}}\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{x-\left( x+h \right)}{x\left( x+h \right)}}{h}.
f(x)=limh0xxhxh(x+h)\Rightarrow {{f}^{'}}\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{x-x-h}{xh\left( x+h \right)}.
f(x)=limh0hxh(x+h)\Rightarrow {{f}^{'}}\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{-h}{xh\left( x+h \right)}.
f(x)=limh01x(x+h)\Rightarrow {{f}^{'}}\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{-1}{x\left( x+h \right)} ---(3).
We know that limxaf(x)=f(a)\displaystyle \lim_{x\to a}f\left( x \right)=f\left( a \right). Let us use this result in equation (3).
f(x)=1x(x+0)\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{-1}{x\left( x+0 \right)}.
f(x)=1x(x)\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{-1}{x\left( x \right)}.
f(x)=1x2\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{-1}{{{x}^{2}}}.
So, we have found the derivative of the function 1x\dfrac{1}{x} using difference quotient method as 1x2\dfrac{-1}{{{x}^{2}}}.
\therefore The derivative of the function 1x\dfrac{1}{x} using a difference quotient method is 1x2\dfrac{-1}{{{x}^{2}}}.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. Whenever we get this type of problem, we first try to recall the required definition to get the required answer. Similarly, we can expect problems to find the derivative of the function sinx\sin x using the difference quotient formula.