Let us assume that:
y=ln(1+x2)
And:
v=tan−1(x)
Derivative of y with respect to x is:
dxdy=dxd(ln(1+x2))=1+x21dxd(1+x2)
=1+x21[dxd(1)+dxd(x2)]
=1+x21(0+2x) [ ∵dxd(xn)=nxn−1anddxd(c)=0 ( Where c is a constant ) ]
=1+x22x
Derivative of v with respect to x is:
dxdv=dxd(tan−1(x))=1+x21 [ ∵dxd(tan−1(x))=1+x21 ]
Now:
dvdy=dxdvdxdy=1+x211+x22x=2x
Hence:
The derivative of ln(1+x2) with respect to tan−1(x) or derivative of y with respect to v is:
dvdy=2x
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