Question

Use logarithmic differentiation to differentiate the following function. f(x) = (x + 4)5(5x - 434 f'(x)=0

Answer 1

`\begin{gathered} y=(x+4)^5(5x-4)^4 \\ by\text{ applying logarithm in both sodes, one has} \\ \ln y=\ln ((x+4)^5(5x-4)^4) \\ \ln y=\ln (x+4)^5+\ln (5x-4)^4 \\ \ln y=5\ln (x+4)+4\ln (5x-4) \\ \end{gathered}`
`\begin{gathered} \text{Now, we w}il\text{ use implicit differentiation:} \\ (1)/(y)(dy)/(dx)=5(1)/(x+4)+4(5)/(5x-4) \\ (1)/(y)(dy)/(dx)=5((1)/(x+4)+(4)/(5x-4)) \\ (1)/(y)(dy)/(dx)=5((5x-4+4(x+4))/((x+4)(5x-4))) \end{gathered}`
`\begin{gathered} (1)/(y)(dy)/(dx)=5((5x-4+4x+16))/((x+4)(5x-4))) \\ (1)/(y)(dy)/(dx)=5((9x+12)/((x+4)(5x-4))) \\ (1)/(y)(dy)/(dx)=15((3x+4)/((x+4)(5x-4))) \\ \text{hence} \\ (dy)/(dx)=15y((3x+4)/((x+4)(5x-4))) \end{gathered}`
`\begin{gathered} by\text{ substituying y, one has} \\ (dy)/(dx)=15(x+4)^5(5x-4)^4((3x+4)/((x+4)(5x-4))) \\ \text{HENCE, THE RESULT IS:} \\ (dy)/(dx)=15(x+4)^4(5x-4)^3(3x+4) \\ OR \\ f(x)^(\prime)=15(x+4)^4(5x-4)^3(3x+4) \end{gathered}`