Question

Differentiate.
y = 6x/(9 − tan x)
Any help would be greatly appreciated!

Answer 1

\[
\left( \frac{f}{g} \right )' = \frac{gf'-fg'}{g^2}
\]

Answer 2

I got an answer but it says its wrong:/ I got (6(9-tans-6xsec^2x))/(9-tans)^2.
Can you tell what I did wrong?

Answer 3

@aum

Answer 4

\[
\left( \frac{6x}{9-\tan(x)} \right )' = \frac{(9-\tan(x))(6x)'-(6x)(9-\tan(x))'}{(9-\tan(x))^2}
= \\
\frac{6(9-\tan(x)) - 6x(-\sec^2(x))}{(9-\tan(x))^2} =
\frac{6(9-\tan(x) + x\sec^2(x))}{(9-\tan(x))^2}
\]

Answer 5

THANK YOU SO MUCH @aum

Answer 6

You are welcome.