Question

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

\(f\left( x \right) = \frac{x}{{{\bf{1}} - {\bf{4}}x}}\)

Short answer

The value of \(f'\left( a \right)\) is \(\frac{1}{{{{\left( {1 - 4a} \right)}^2}}}\).

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:

\(\begin{aligned}f'\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{a + h}}{{1 - 4\left( {a + h} \right)}} - \frac{a}{{1 - 4a}}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left( {a + h} \right)\left( {1 - 4a} \right) - a\left( {1 - 4\left( {a + h} \right)} \right)}}{{\left( {1 - 4a} \right)\left( {1 - 4\left( {a + h} \right)} \right)}}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{a - 4{a^2} + h - 4ah - a + 4{a^2} + 4ah}}{{h\left( {1 - 4a} \right)\left( {1 - 4\left( {a + h} \right)} \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{h}{{h\left( {1 - 4a} \right)\left( {1 - 4\left( {a + h} \right)} \right)}}\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( a \right) &= \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {1 - 4a} \right)\left( {1 - 4\left( {a + h} \right)} \right)}}\\ &= \frac{1}{{\left( {1 - 4\left( {a + 0} \right)} \right)\left( {1 - 4a} \right)}}\\ &= \frac{1}{{\left( {1 - 4a} \right)\left( {1 - 4a} \right)}}\\ &= \frac{1}{{{{\left( {1 - 4a} \right)}^2}}}\end{aligned}\)

Thus, the value of \(f'\left( a \right)\) is \(\frac{1}{{{{\left( {1 - 4a} \right)}^2}}}\).