Question

Use the Chain Rule to differentiate the function. You may need to apply the rule more than once. f(x) = (4x3 - (8x + 9)2) O f'(x) = 7[4x3 - (8x + 9)216[12x2 - 16(8x + 9)] O f'(x) = 7[4x3 - (8x + 9)21][12x1 - 16(8x + 9)] O f'(x) = 7[4x3 - (8x + 9)216[12x2 -2(8x + 9)] O f'(x) = 7[4x3 - (8x + 9)21][12x1 - 2(8x + 9)]

Answer 1

We will express this function as 3 function nested:

$\begin{gathered} f(x)=\lbrack g(x)\rbrack^7 \\ g(x)=4x^3-\lbrack h(x)\rbrack^2 \\ h(x)=8x+9 \end{gathered}$

Then, the chain rule can be written as:

$\frac{df}{dx}=\frac{df}{dg}\cdot\frac{dg}{dh}\cdot\frac{dh}{dx}$

In this case, as g(x) have two terms, and only one of them is referred to h(x), we can solve itlike that:

$\frac{dh}{dx}=8\cdot\frac{d(x)}{dx}+9\frac{d(1)}{dx}=8\cdot1+0=8$

Then, g'(x) is:

$\begin{gathered} \frac{dg}{dx}=4(3x^2)-2h(x)\cdot\frac{dh}{dx}=12x^2-2(8x+9)\cdot8 \\ \frac{dg}{dx}=12x^2-16\cdot8x-16\cdot9 \\ \frac{dg}{dx}=12x^2-128x-144 \end{gathered}$$\begin{gathered} \frac{df}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}=(7\cdot g(x)^6)\cdot(12x^2-128x-144) \\ \frac{df}{dx}=7\cdot\lbrack4x^3-(8x+9)^2\rbrack6\cdot(12x^2-128x-144) \\ \frac{df}{dx}=42\cdot\lbrack4x^3-(8x+9)^2\rbrack\cdot4\cdot(3x^2-32x-36) \\ \frac{df}{dx}=168\lbrack4x^3-(8x+9)^2\rbrack\cdot(3x^2-32x-36) \end{gathered}$