[tex]\ln(4x+y)=2x-3[/tex]
Differentiate both sides wrt [tex]x[/tex]:
[tex]\dfrac{\mathrm d(\ln(4x+y))}{\mathrm dx}=\dfrac{\mathrm d(2x-3)}{\mathrm dx}[/tex]
By the chain rule, we get
[tex]\dfrac1{4x+y}\dfrac{\mathrm d(4x+y)}{\mathrm dx}=2[/tex]
[tex]\dfrac{4+\frac{\mathrm dy}{\mathrm dx}}{4x+y}=2[/tex]
Solve for [tex]\frac{\mathrm dy}{\mathrm dx}[/tex]:
[tex]4+\dfrac{\mathrm dy}{\mathrm dx}=8x+2y[/tex]
[tex]\boxed{\dfrac{\mathrm dy}{\mathrm dx}=8x+2y-4}[/tex]
