Question
Differentiate y=log(8x^2-3x+9)
Answers
Jai
y = log(8x^2-3x+9)
Note that derivative of log(x) = 1/(x*ln(10))
Thus,
dy/dx
= ( 1/(8x^2-3x+9)*ln(10) ) * (16x - 3)
= (16x - 3) / ((8x^2 - 3x + 9)*ln(10))
Hope this helps~ :3
Note that derivative of log(x) = 1/(x*ln(10))
Thus,
dy/dx
= ( 1/(8x^2-3x+9)*ln(10) ) * (16x - 3)
= (16x - 3) / ((8x^2 - 3x + 9)*ln(10))
Hope this helps~ :3