Question

3-34: Differentiate the function

24. \(y = \frac{{\sqrt x + x}}{{{x^2}}}\)

Short answer

The derivative of the function is \(f'\left( x \right) = - \frac{3}{{2{x^{\frac{5}{2}}}}} - \frac{1}{{{x^2}}}\).

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

Rewrite the function as the power of \(x\) as shown below:

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

Differentiate the function as shown below:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {{x^{ - \frac{3}{2}}} + {x^{ - 1}}} \right)\\ &= \frac{d}{{dx}}\left( {{x^{ - \frac{3}{2}}}} \right) + \frac{d}{{dx}}\left( {{x^{ - 1}}} \right)\\ &= - \frac{3}{2}\left( {{x^{ - \frac{5}{2}}}} \right) + \left( { - 1} \right){x^{ - 2}}\\ &= - \frac{3}{{2{x^{\frac{5}{2}}}}} - \frac{1}{{{x^2}}}\end{aligned}\)

Thus, the derivative of the function is \(f'\left( x \right) = - \frac{3}{{2{x^{\frac{5}{2}}}}} - \frac{1}{{{x^2}}}\).