I Differentiated it
d tan(cos-1x)d x
hint given is
=>tanθ=1√1+x2
which I think is not possible in any way
Answer given inthe book is
= -3x(1+x2)-3/2
the way i did the sum is as follows
cos-1x=θ
=>tanθ=√1-x2x
.'.
= x d√1-x2dx - √1-x2dxdxx2
= -1x2√1-x2 ................(ans)
P.S. Suggest the correct ans
71You didn't looked at the right side of the expression you had differentiated.
I.e., tanθ = √1-x2x ........I
When You differentiate, the right side is as you have shown but,
sec2θ . dθdt = expression you have written ....
Now, Find sec2θ . dθdt from relation I
