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 \(dy/dx\).

2.\(y = \sqrt {{x^3} + 2} \)

Short answer

Derivative of given function is \(y' = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 2} }}\).

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 \(y = \sqrt {{x^3} + 2} \)

Let \(\left( {{x^3} + 2} \right) = u\). Then,

\(\begin{array}{l}\left( {{x^3} + 2} \right) = u\\g\left( x \right) = {x^3} + 2\\f\left( {g\left( u \right)} \right) = \sqrt {\left( {{x^3} + 2} \right)} \end{array}\)

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.

\(\begin{aligned}\frac{{du}}{{dx}} &= \frac{d}{{dx}}\left( {{x^3} + 2} \right)\\u' &= 3{x^2}\end{aligned}\)

\(\begin{aligned}{c}\frac{{df}}{{du}} &= \frac{d}{{du}}\sqrt {\left( u \right)} \\f' &= \frac{1}{{2\sqrt {\left( u \right)} }}\\f' &= \frac{1}{{2\sqrt {{x^3} + 2} }}\end{aligned}\)

Chain rule of differentiation

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

Find the derivative of y with respect to x as:

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{df}}{{du}} \cdot \frac{{du}}{{dx}}\\y' &= \frac{1}{{2\sqrt {{x^3} + 2} }}3{x^2}\\y' &= \frac{{3{x^2}}}{{2\sqrt {{x^3} + 2} }}\end{aligned}\)

Hence, the differentiation of the function is \(y' = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 2} }}\).