Question

3-34: Differentiate the function

23. \(f\left( x \right) = \frac{{3{x^2} + {x^3}}}{x}\)

Short answer

The derivative of the function is \(f'\left( x \right) = 3 + 2x\).

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;

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

When \(n\) is any real number, then;

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

Differentiate the function

Simplify the function as shown below:

\(\begin{aligned}f\left( x \right) &= \frac{{3{x^2} + {x^3}}}{x}\\ &= \frac{{3{x^2}}}{x} + \frac{{{x^3}}}{x}\\ &= 3x + {x^2}\end{aligned}\)

Differentiate the function as shown below:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {3x + {x^2}} \right)\\ &= 3\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {{x^2}} \right)\\ &= 3\left( 1 \right) + \left( 2 \right){x^{2 - 1}}\\ &= 3 + 2x\end{aligned}\)

Thus, the derivative of the function is \(f'\left( x \right) = 3 + 2x\).