Question

Find \(f'\left( x \right)\). Check your answer is reasonable by comparing the graphs of \(f\), \(f'\).

80. \(f\left( x \right) = \arctan \left( {{x^2} - x} \right)\)

Short answer

It is clear from the graph that the curve of \(f'\) crosses the \(x\)-axis where the graph of \(f\) has a horizontal tangent. The graph of \(f\) is decreasing where \(f'\) is negative and \(f\) is increasing where \(f'\) is positive.

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the function:

Differentiate \(f\left( x \right)\) with respect to \(x\) as follows:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {\arctan \left( {{x^2} - x} \right)} \right)\\ &= \frac{1}{{1 + {{\left( {{x^2} - x} \right)}^2}}} \cdot \frac{d}{{dx}}\left( {{x^2} - x} \right)\\ &= \frac{1}{{1 + {{\left( {{x^2} - x} \right)}^2}}} \cdot \left( {2x - 1} \right)\\ &= \frac{{2x - 1}}{{1 + {{\left( {{x^2} - x} \right)}^2}}}\end{aligned}\)

Hence, \(f'\left( x \right) = \frac{{2x - 1}}{{1 + {{\left( {{x^2} - x} \right)}^2}}}\).

Check the answer visually

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

To check the answer visually draw the graph of the function\(f\left( x \right) = \arctan \left( {{x^2} - x} \right)\), \(f'\left( x \right) = \frac{{2x - 1}}{{1 + {{\left( {{x^2} - x} \right)}^2}}}\) using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\arctan \left( {{x^2} - x} \right)\)in the\({Y_1}\)tab.
2. Select the “STAT PLOT” and enter the equation\(\frac{{2x - 1}}{{1 + {{\left( {{x^2} - x} \right)}^2}}}\)in the\({Y_2}\)tab.
3. Enter the “GRAPH” button in the graphing calculator.

Visualization of graph of the function\(f\left( x \right) = \arctan \left( {{x^2} - x} \right)\), \(f'\left( x \right) = \frac{{2x - 1}}{{1 + {{\left( {{x^2} - x} \right)}^2}}}\) is shown below:

It is clear from the graph that the curve of \(f'\) crosses the \(x\)-axis where the graph of \(f\) has a horizontal tangent.

The graph of \(f\) is decreasing where \(f'\) is negative and \(f\) is increasing where \(f'\) is positive.