Differentiate x^sinx, x>0 w.r.t. x.
Answers!
Shruti Shibu
Sept. 10, 2019, 9:02 a.m.
y=x^sinx
log y= log x^sinx
log y= sinx logx
1/y(dy/dx) = cosx logx + (sinx)/x
dy/dx= y[cosx logx = 1]
dy/dx = x^sinx (cosx log x + 1)
Comment
Answer the question...
Shruti Shibu
y=x^sinx
log y= log x^sinx
log y= sinx logx
1/y(dy/dx) = cosx logx + (sinx)/x
dy/dx= y[cosx logx = 1]
dy/dx = x^sinx (cosx log x + 1)
Comment
Go ad-free with Teachoo Black
Adblock detected
Skip every ad. Keep every answer.
Teachoo Black gives you a clean, uninterrupted study experience on every page.
Go ad-free →
A second ad-free Teachoo reminder
One last thing
This is what ad-free feels like.
No banners. No interruptions. Just the lesson.
Go ad-free →Please login to view more pages. It's free :)
Teachoo gives you a better experience when you're logged in. Please login :)