Question

Differentiate the function.

\(j\left( x \right) = {x^{2.4}} + {e^{2.4}}\)

Short answer

Derivative of the above function is \(2.4{x^{1.4}}\).

Step-by-step solution

Precise Definition of differentiation

A derivative is a function that defines the rate of change of a variable. In geometry,the derivative is a function that can be interpreted as the slope of the graph of the function.

Sum rule of differentiation

Use the sum rule of differentiation and apply it here

\(\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\)

\(\frac{d}{{dx}}\left( {j\left( x \right)} \right) = \frac{d}{{dx}}\left( {{x^{2.4}}} \right) + \frac{d}{{dx}}\left( {{e^{2.4}}} \right)\)

Power rule of differentiation

The power rule of differentiation implies is \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\), where \(n\) is any real number.

Apply power rule and simplify the above function.

\(\frac{d}{{dx}}\left( {j\left( x \right)} \right) = 2.4\left( {{x^{2.4 - 1}}} \right) + \frac{d}{{dx}}\left( {{e^{2.4}}} \right)\)

Constant rule of differentiation

Here \({e^{2.4}}\) is the constant term so the derivative of this one is zero.

\(\begin{aligned}j'\left( x \right) &= 2.4\left( {{x^{1.4}}} \right) + 0\\ &= 2.4{x^{1.4}}\end{aligned}\)

Hence, the differentiation of the function is \(j'\left( x \right) = 2.4{x^{1.4}}\).