Question

find the derivative:
y=ln(e^-t tant)

Answer 1

Get ready to use the chain rule and the product rule.
\[y'=\frac{1}{e^{-t} \tan t} \times \left[ e^{-t} \tan t \right]'\]
How do you differentiate that second part?

Answer 2

I don't know ..

How?

Answer 3

Product rule:
\[\left[ f(x) \cdot g(x) \right]' = f'(x) \cdot g(x) + f(x) \cdot g'(x) \]

Answer 4

what derivative 1/e^-t tant
and
(e^-t tant)'

Answer 5

I'm want just first differentiate..
thanks

Answer 6

Can we simplify?