Question

3-34: Differentiate the function

20. \(F\left( t \right) = {\left( {2t - 3} \right)^2}\)

Short answer

The derivative of the function is \(F'\left( t \right) = 8t - 12\).

Step-by-step solution

 Differentiation Rule

When \(c\) is a constant and \(f\) is a differentiable function, then;

\(\frac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\frac{d}{{dx}}f\left( x \right)\)

\(\frac{d}{{dx}}\left( x \right) = 1\)

The derivative of the constant function is;

\(\frac{d}{{dx}}\left( c \right) = 0\)

When \(f\) and \(g\) are both differentiable, then;

\(\begin{array}{l}\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) &=& \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\\\frac{d}{{dx}}\left( {f\left( x \right) - g\left( x \right)} \right) &=& \frac{d}{{dx}}f\left( x \right) - \frac{d}{{dx}}g\left( x \right)\end{array}\)

Differentiate the function

Expand the function as shown below:

\(\begin{aligned}F\left( t \right) &= {\left( {2t - 3} \right)^2}\\ &= 4{t^2} - 12t + 9\end{aligned}\)

Differentiate the function as shown below:

\(\begin{aligned}F'\left( t \right) &= \frac{d}{{dt}}\left( {4{t^2} - 12t + 9} \right)\\ &= 4\frac{d}{{dt}}\left( {{t^2}} \right) - 12\frac{d}{{dt}}\left( t \right) + \frac{d}{{dt}}\left( 9 \right)\\ &= 4\left( 2 \right){t^{2 - 1}} - 12\left( 1 \right) + 0\\ &= 8t - 12\end{aligned}\)

Thus, the derivative of the function is \(F'\left( t \right) = 8t - 12\).