Question

Find the first and second derivatives of the function.

49. \(f\left( x \right) = 0.001{x^5} - 0.02{x^3}\)

Short answer

The first derivative is \(f'\left( x \right) = 0.005{x^4} - 0.06{x^2}\)and the second derivative is \(f''\left( x \right) = 0.02{x^3} - 0.12x\).

Step-by-step solution

Find the first derivative

The given function is \(f\left( x \right) = 0.001{x^5} - 0.02{x^3}\).

Use difference rule of differentiation which implies that \(\frac{d}{{dx}}\left( {f\left( x \right) - g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) - \frac{d}{{dx}}g\left( x \right)\) and power rule of differentiation which implies that \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\), where \(n\) is any real number.

On differentiating this function with respect to \(x\) by using difference and power rule, the following expression is obtained:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}\left( {0.001{x^5} - 0.02{x^3}} \right)\\ &= \frac{d}{{dx}}\left( {0.001{x^5}} \right) - \frac{d}{{dx}}\left( {0.02{x^3}} \right)\\ &= 0.001\left( {5{x^4}} \right) - 0.02\left( {3{x^2}} \right)\\ &= 0.005{x^4} - 0.06{x^2}\end{aligned}\)

Thus, the first derivative is \(f'\left( x \right) = 0.005{x^4} - 0.06{x^2}\).

Find the second derivative

The first derivative of the function is \(f'\left( x \right) = 0.005{x^4} - 0.06{x^2}\). On differentiating it again, with respect to \(x\) axis by using difference and power rule, we obtain the second derivative, as follows:

\(\begin{aligned}f''\left( x \right) &= \frac{d}{{dx}}\left( {0.005{x^4} - 0.06{x^2}} \right)\\ &= \frac{d}{{dx}}\left( {0.005{x^4}} \right) - \frac{d}{{dx}}\left( {0.06{x^2}} \right)\\ &= 0.005\left( {4{x^3}} \right) - 0.06\left( {2x} \right)\\ &= 0.02{x^3} - 0.12x\end{aligned}\)

Thus, the second derivative is \(f''\left( x \right) = 0.02{x^3} - 0.12x\).