Question

Using the first principle find the deriavative:1/x^2+1

Answer 1

To find the derivative of the function \( f(x) = \frac{1}{x^2} + 1 \) using the first principle of derivatives, we start with the definition of the derivative:

\[
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
\]

First, we need to compute \( f(x + h) \):

\[
f(x + h) = \frac{1}{(x + h)^2} + 1
\]

Now we can substitute this back into our formula for the derivative:

\[
f' (x) = \lim_{h \to 0} \frac{ \left( \frac{1}{(x + h)^2} + 1 \right) - \left( \frac{1}{x^2} + 1 \right)}{h}
\]

Simplifying the expression in the numerator:

\[
f' (x) = \lim_{h \to 0} \frac{ \frac{1}{(x + h)^2} - \frac{1}{x^2}}{h}
\]

To combine the fractions in the numerator, we find a common denominator:

\[
\frac{1}{(x + h)^2} - \frac{1}{x^2} = \frac{x^2 - (x + h)^2}{x^2 (x + h)^2}
\]

Now calculate \( (x + h)^2 \):

\[
(x + h)^2 = x^2 + 2xh + h^2
\]

So we have:

\[
x^2 - (x + h)^2 = x^2 - (x^2 + 2xh + h^2) = -2xh - h^2
\]

Thus,

\[
\frac{1}{(x + h)^2} - \frac{1}{x^2} = \frac{-2xh - h^2}{x^2 (x + h)^2}
\]

Substituting this back into the derivative:

\[
f' (x) = \lim_{h \to 0} \frac{-2xh - h^2}{h \cdot x^2 (x + h)^2}
\]

We can simplify the fraction by canceling \( h \) (note \( h \neq 0 \)):

\[
f' (x) = \lim_{h \to 0} \frac{-2x - h}{x^2 (x + h)^2}
\]

Now, taking the limit as \( h \to 0 \):

\[
f' (x) = \frac{-2x}{x^2 (x)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}
\]

Thus, the derivative of \( f(x) = \frac{1}{x^2} + 1 \) is:

\[
f'(x) = -\frac{2}{x^3}
\]