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Last updated date: 17th Apr 2024
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Answer
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Hint- Try to solve using the definition of derivative i.e. \[f\prime \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]

Let \[f\left( x \right) = x\]
We need to find the derivative of \[f\left( x \right)\]at\[{\text{x = 1}}\].
\[{\text{i}}{\text{.e}}{\text{. }}f\prime \left( 1 \right)\]
We know that,
\[f\prime \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}{\text{ }} \ldots \left( 1 \right)\]
Here, \[f\left( x \right) = x\]
So, \[{\text{ }}f\left( {x + h} \right) = x + h\]
Putting these values in equation $\left( 1 \right)$, we get
\[
  f\prime \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {x + h} \right) - x}}{h} \\
   = \mathop {\lim }\limits_{h \to 0} \dfrac{{x + h - x}}{h} \\
   = \mathop {\lim }\limits_{h \to 0} {\text{ }}\dfrac{h}{h} \\
   = \mathop {\lim }\limits_{h \to 0} {\text{ }}1 \\
   = 1 \\
\]
Hence, \[f\prime \left( x \right) = 1\]
Putting $x = 1$, we get
\[f\prime \left( 1 \right) = 1\]
Hence, the derivative of $x$ at $x=1$ is $1$.

Note- In order to calculate the derivative of a certain function, we assume that function to be $f\left( x \right)$ and apply the formula of the derivative and later put the value of $x$ for which the derivative is asked to calculate.