If x2+y2=aetan-1yx, a>0,y(0)≠0 then f''(0) is:
aee-π/2
aeπ/2
-2e-π/2a
aeeπ/2
Explanation for the correct option:
Second order derivative:
In x2+y2=aetan-1yx, by taking log both sides, we get
logx2+y2=logaetan-1yx⇒12logx2+y2=loga+logetan-1yx⇒12logx2+y2=loga+tan-1yx
By differentiating both sides w.r.t x, we get
12×1x2+y2×2x+2yy'=11+yx2×xy'-yx2⇒x+yy'x2+y2=xy'-yx2+y2⇒x+yy'=xy'-y..........(1)⇒y'=x+yx-y⇒y'0=-1
By differentiating 1 w.r.t x, we get
1+yy''+y'2=xy''+y'-y'⇒y''=1+y'2x-y⇒y''0=1+1aeπ/2[∵y0=aeπ/2]⇒y''0=2e-π/2a
Hence, option C is correct.