wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the derivative of f(x)=xcosx1+x2.

Open in App
Solution

wehavef(x)=xcosx1+x2Putx=tantdx=sec2dtf(tant)=tantcos(tant)sectf(x)dx=f(tant)sec2tdt=d(sintcos(tant))=(costcos(tant)sintsin(tant)sec2t)dt=[cos3tcos(tant)sintsin(tant)]dxf(x)dx=[cos3tcos(tant)sintsin(tant)]dx=(11+x2)3cosxx1+x2sinxdxddx[f(x)]=⎢ ⎢ ⎢ ⎢cosx(1+x2)32xsinx(1+x2)12⎥ ⎥ ⎥ ⎥

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Basic Theorems in Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon