Question

derivative y=(x+cosx)^sinx

Answer 1

since the exponent is the function of 'x', you need to use logarithmic differentiation

so, start by taking log on both sides,
what do u get ?

Answer 2

lny = ln[(x+cosx)^sinx

i dont know what to do after this

Answer 3

use the log property for right side
\(\Large \log a^b = b \log a\)

Answer 4

lny = sinxln(x+cosx) ?

Answer 5

yes
now you can differentiate both sides!

chain rule + product rule

try to proceed ?

Answer 6

cosx * ln(x+cosx) + sinx * 1/(x+cosx) * 1- sinx?

Answer 7

yes thats correct.
and on left side, its 1/y dy/dx

Answer 8

so i multiply the right by y right?

Answer 9

correct
and plug back in
y= (x+cosx)^sinx on right side
and that would be it

Answer 10

ok thank you man! i appreciat ethe help!

Answer 11

welcome ^_^