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Question

Question: Find the derivative of x at x = 1....

Find the derivative of x at x = 1.

Explanation

Solution

Hint- Try to solve using the definition of derivative i.e. f(x)=limh0f(x+h)f(x)hf\prime \left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}

Let f(x)=xf\left( x \right) = x
We need to find the derivative of f(x)f\left( x \right)atx = 1{\text{x = 1}}.
i.e. f(1){\text{i}}{\text{.e}}{\text{. }}f\prime \left( 1 \right)
We know that,
f(x)=limh0f(x+h)f(x)h (1)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(x)=xf\left( x \right) = x
So,  f(x+h)=x+h{\text{ }}f\left( {x + h} \right) = x + h
Putting these values in equation (1)\left( 1 \right), we get

f(x)=limh0(x+h)xh =limh0x+hxh =limh0 hh =limh0 1 =1  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(x)=1f\prime \left( x \right) = 1
Putting x=1x = 1, we get
f(1)=1f\prime \left( 1 \right) = 1
Hence, the derivative of xx at x=1x=1 is 11.

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