Question

3-34: Differentiate the function

21. \(y = 3{e^x} + \frac{4}{{\sqrt[3]{x}}}\)

Short answer

The derivative of the function is \(f'\left( x \right) = 3{e^x} - \frac{4}{{3x\sqrt[3]{x}}}\).

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;

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

The derivative of the natural exponential function is;

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

Differentiate the function

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

\(\begin{aligned}y &= f\left( x \right)\\ &= 3{e^x} + \frac{4}{{\sqrt[3]{x}}}\\ &= 3{e^x} + 4{x^{ - \frac{1}{3}}}\end{aligned}\)

Differentiate the function as shown below:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {3{e^x} + 4{x^{ - \frac{1}{3}}}} \right)\\ &= 3\frac{d}{{dx}}\left( {{e^x}} \right) + 4\frac{d}{{dx}}\left( {{x^{ - \frac{1}{3}}}} \right)\\ &= 3\left( {{e^x}} \right) + 4\left( { - \frac{1}{3}} \right){x^{ - \frac{1}{3} - 1}}\\ &= 3{e^x} - \frac{4}{3}{x^{ - \frac{4}{3}}}\\ &= 3{e^x} - \frac{4}{{3x\sqrt[3]{x}}}\end{aligned}\)

Thus, the derivative of the function is \(f'\left( x \right) = 3{e^x} - \frac{4}{{3x\sqrt[3]{x}}}\).