Question 12.2: If y = (48 + 20x^−1 + 4x + 0.3x²)^4 , what is dy/dx?...

Let

z = 48 + 20x^{-1} + 4x + 0.3x²         (1)

and so

\frac{dz}{dx} = −20x^{-2} + 4 + 0.6x         (2)

Substituting (1) into the function given in the question

y = z^{4}

and so

\frac{dy}{dz} = 4z³        (3)

Therefore, using the chain rule and substituting (2) and (3)

\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx}
          = 4z³(−20x^{-2} + 4 + 0.6x)
          = 4(48 + 20x^{-1} + 4x + 0.3x²)³(−20x^{-2} + 4 + 0.6x)