Question

19-20 Use Definition 4 to find\(f'\left( a \right)\)at the given number\(a\).

19.\(f\left( x \right) = \sqrt {4x + 1} ,\,\,a = 6\)

Short answer

The required answer is \(f'\left( 6 \right) = \frac{2}{5}\).

Step-by-step solution

Derivative of a function using limit

For a given function \(f\) we can find the derivative of the function by using limit. Which is given by \(\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = f'\left( x \right)\).

Finding the derivative of a function at a given point

The function is given by \(f\left( x \right) = \sqrt {4x + 1} \).

Now the derivative of the function at \(x = 6\) will be:

\(\begin{aligned}f'\left( 6 \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {6 + h} \right) - f\left( 6 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {4\left( {6 + h} \right) + 1} - \sqrt {6 \times 4 + 1} }}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {25 + 4h} - 5}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {25 + 4h} - 5} \right)\left( {\sqrt {25 + 4h} + 5} \right)}}{{h\left( {\sqrt {25 + 4h} + 5} \right)}}\end{aligned}\)

Further solving we get,

\(\begin{aligned}f'\left( 6 \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{25 + 4h - 25}}{{h\left( {\sqrt {25 + 4h} + 5} \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{4h}}{{h\left( {\sqrt {25 + 4h} + 5} \right)}}\\ &= \frac{4}{{\sqrt {25} + 5}}\\ &= \frac{4}{{5 + 5}}\\ &= \frac{2}{5}\end{aligned}\)

Hence, the required value is \(f'\left( 6 \right) = \frac{2}{5}\).