Question

1-6: Write the composite function in the form of \(f\left( {g\left( x \right)} \right)\).

(Identify the inner function \(u = g\left( x \right)\)and outer function\(y = f\left( u \right)\).) Then find the derivative \(dy/dx\).

1.\(y = {\left( {5 - {x^4}} \right)^3}\)

Short answer

Derivative of given function is \( - 12{x^3}{\left( {5 - {x^4}} \right)^2}\)

Step-by-step solution

Chain rule of differentiation

Use the rule of differentiation which can apply on this question:

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

\(\)

Now we can apply on \(y = {\left( {5 - {x^4}} \right)^3}\).

Let \(\left( {5 - {x^4}} \right) = u\). Then,

\(\begin{aligned}\left( {5 - {x^4}} \right) &= u\\g\left( x \right) &= 5 - {x^4}\\f\left( {g\left( u \right)} \right) &= {\left( {5 - {x^4}} \right)^3}\end{aligned}\)

Power rule of differentiation

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

Apply power rule and simplify.

\(\begin{aligned}\frac{{du}}{{dx}} &= \frac{d}{{dx}}\left( {5 - {x^4}} \right)\\u' &= 0 - 4{x^3}\\u' &= - 4{x^3}\end{aligned}\)

\(\begin{aligned}\frac{{df}}{{du}} &= \frac{d}{{du}}\left( {{u^3}} \right)\\ &= 3{u^2}\\ &= 3{\left( {5 - {x^4}} \right)^2}\end{aligned}\)

Chain rule of differentiation

The chain rule of differentiation is that \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\).

Find \(\frac{{dy}}{{dx}}\).

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{df}}{{du}} \cdot \frac{{du}}{{dx}}\\y' &= 3{\left( {5 - {x^4}} \right)^2}\left( { - 4{x^3}} \right)\\y' &= - 12{x^3}{\left( {5 - {x^4}} \right)^2}\end{aligned}\)

Hence, the differentiation of the function is \(y' = - 12{x^3}{\left( {5 - {x^4}} \right)^2}\).