Question

Find \(f'\left( a \right)\).

\(f\left( t \right) = \frac{{\bf{1}}}{{{t^{\bf{2}}} + {\bf{1}}}}\)

Short answer

The value of \(f'\left( a \right)\) is \( - \frac{{2a}}{{{{\left( {{a^2} + 1} \right)}^2}}}\).

Step-by-step solution

Use the definition 4

The derivative of a function at a number \(a\), denoted by \(f'\left( a \right)\) can be obtained using the formula given below:

\(f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\)

The derivative of the given function

Substitute the values in the above formula and solve it as follows:

\(\begin{aligned}f'\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{{{\left( {a + h} \right)}^2} + 1}} - \frac{1}{{{a^2} + 1}}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left( {{a^2} + 1} \right) - \left( {{{\left( {a + h} \right)}^2} + 1} \right)}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)}}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{a^2} + 1} \right) - \left( {{{\left( {a + h} \right)}^2} + 1} \right)}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)\left( h \right)}}\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{a^2} + 1} \right) - \left( {{a^2} + {h^2} + 2ah + 1} \right)}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)\left( h \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} - 2ah}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)\left( h \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {h - 2a} \right)}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)\left( h \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {h - 2a} \right)}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)}}\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {h - 2a} \right)}}{{\left( {{{\left( {a + h} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)}}\\ &= \frac{{0 - 2a}}{{\left( {{{\left( {a + 0} \right)}^2} + 1} \right)\left( {{a^2} + 1} \right)}}\\ &= - \frac{{2a}}{{\left( {{a^2} + 1} \right)\left( {{a^2} + 1} \right)}}\\ &= - \frac{{2a}}{{{{\left( {{a^2} + 1} \right)}^2}}}\end{aligned}\)

Thus, the value of \(f'\left( a \right)\) is \( - \frac{{2a}}{{{{\left( {{a^2} + 1} \right)}^2}}}\).