CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find dydx of y=tanxcotx+cotxtanx.


Open in App
Solution

Compute the required derivative:

Given: y=tanxcotx+cotxtanx.

y=elogtanxcotx+elogcotxtanxy=ecotxlogtanx+etanxlogcotx

Taking dydxon both sides,

dydx=ecotxlogtanxcotxsec2xtanx-csc2xlogtanx+etanxlogcotxtanx-csc2xcotx+sec2xlogcotxddxelogxy=eylogx×d(ylogx)dxdydx=tanxcotx(csc2x-csc2xlogtanx)+cotxtanx(-sec2x+sec2xlogcotx)dydx=tanxcotxcsc2x1-logtanx+cotxtanxsec2x-1+logcotx

Hence, dydx=tanxcotxcsc2x1-logtanx+cotxtanxsec2x-1+logcotx.


flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon