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:

`(df)/(dx)=(df)/(dg)\cdot(dg)/(dh)\cdot(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:

`(dh)/(dx)=8\cdot(d(x))/(dx)+9(d(1))/(dx)=8\cdot1+0=8`

Then, g'(x) is:

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