f(x) =lin(x+linx)
Given,f(x)=ln(x+lnx)Given,\\f(x)=ln(x+lnx)Given,f(x)=ln(x+lnx)
and we know that ddxlnx=1x\dfrac{d}{dx}lnx=\dfrac{1}{x}dxdlnx=x1
So,
f′(x)=ddxln(x+lnx) f′(x)=1(x+lnx)ddx[x+lnx]=1(1+lnx)⋅[1+1x]=x+1x(1+lnx)f'(x)=\dfrac{d}{dx}ln(x+lnx)\\\ \\f'(x)=\dfrac{1}{(x+lnx)}\dfrac{d}{dx}[x+lnx]=\dfrac{1}{(1+lnx)}\cdot[1+\dfrac{1}{x}]=\dfrac{x+1}{x(1+lnx)}f′(x)=dxdln(x+lnx) f′(x)=(x+lnx)1dxd[x+lnx]=(1+lnx)1⋅[1+x1]=x(1+lnx)x+1
Hence, f′(x)=x+1x⋅(1+lnx)\boxed{f'(x)=\dfrac{x+1}{x\cdot(1+lnx)}}f′(x)=x⋅(1+lnx)x+1
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