Question

F(x) = (8x - 4) ^ 3 * (4x ^ 2 + 7) ^ 4 find f^ prime (x) using logarithmic differentiation

Answer 1

To solve this question we going to need to formulas:

`\begin{gathered} ln(x\cdot y)=ln(x)+ln(y)\text{ \lparen1\rparen} \\ \\ ln(x^y)=yln(x)\text{ }\left(2\right) \end{gathered}`

Now to make a logarithmic differentiation we apply logarithm to the initial equation

`\begin{gathered} f\mleft(x\mright)=(8x-4)^3*(4x^2+7)^4 \\ \\ ln(f(x))=ln((8x-4)^3(4x^2+7)^4) \end{gathered}`

Now we apply the first two formulas

`\begin{gathered} ln(f(x))=ln((8x-4)^3)+ln((4x^2+7)^4) \\ \\ ln(f(x))=3\cdot ln((8x-4))+4\cdot ln((4x^2+7)) \end{gathered}`

Now we make an implicit derivate

`(f^(\prime)(x))/(f\lparen x))=(6)/(2x-1)+(32x)/(4x^2+7)`

Simplify

Now we are almost done but we have that f(x), but we know what is that from the beginning, f(x) = (8x - 4) ^ 3 * (4x ^ 2 + 7) ^ 4

Replacing

`f^(\prime)(x)=((8x-4)^3*(4x^2+7)^4)\cdot(88x^(2)-32x+42)/((2x-1)(4x^(2)+7))`

Answer:

Simplify

`f^(\prime)(x)=64\left(2x-1\right)^2\left(4x^2+7\right)^3\left(88x^2-32x+42\right)`