Derivative of 7^(SinX)... something to do with log/ln?
Calculus 1 - Janelle, Thursday, June 21, 2012 at 2:24am
You have to use the chain rule.
ln(7) * 7^(sin x) * cos (x)
Calculus 1 - Reiny, Thursday, June 21, 2012 at 7:49am
let y = 7^(sinx)
take ln of both sides
ln y = ln (7^sinx)
ln y = sinx (ln7)
now use the rule for ln
y' /y = ln7(cosx)
y' = ln7(cosx) * (7^sinx)