To solve this question we going to need to formulas:
[tex]\begin{gathered} ln(x\cdot y)=ln(x)+ln(y)\text{ \lparen1\rparen} \\ \\ ln(x^y)=yln(x)\text{ }\left(2\right) \end{gathered}[/tex]
Now to make a logarithmic differentiation we apply logarithm to the initial equation
[tex]\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}[/tex]
Now we apply the first two formulas
[tex]\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}[/tex]
Now we make an implicit derivate
[tex]\frac{f^{\prime}(x)}{f\lparen x)}=\frac{6}{2x-1}+\frac{32x}{4x^2+7}[/tex]
Simplify
[tex]\frac{f^{\prime}(x)}{f\operatorname{\lparen}x)}=\frac{88x^2-32x+42}{\left(2x-1\right)\left(4x^2+7\right)}[/tex]
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
[tex]f^{\prime}(x)=((8x-4)^3*(4x^2+7)^4)\cdot\frac{88x^{2}-32x+42}{(2x-1)(4x^{2}+7)}[/tex]
Answer:
Simplify
[tex]f^{\prime}(x)=64\left(2x-1\right)^2\left(4x^2+7\right)^3\left(88x^2-32x+42\right)[/tex]
