Question

Differentiate the following: \[y=log(8x^2-3x+9)\]

Answer 1

\[y'=\frac{ 1 }{ (8x^2-3x+9) }*16x-3\]\]

Answer 2

What is the derivative of log u?
Use that plus the chain rule.

Answer 3

thank you

Answer 4

Correct. Make sure 16x - 3 is in parentheses or it's simply the numerator of the fraction.

Answer 5

word

Answer 6

\(y'=\dfrac{16x-3 }{8x^2-3x+9 } \)

Answer 7

You're welcome.