1.

Given, `y=(ax+b)^(2)` , evaluate ` (dy)/(dx)`.

Answer» Method I :
Substituting (ax+b)=u
Then `" "(du)/(dx)=(d(ax+b))/(dx)=a`
and `" "(dy)/(du)=(d(ax+b)^(2))/(du)=(d(u)^(2))/(du)=2u`
`:. " " (dy)/(dx)=(dy)/(du) xx (du)/(dx)=2u xx a =2 (ax+b)a`
`" " =2a(ax+b)`
Method II :
`y=(ax+b)^(2)`
`=a^(2)x^(2)+b^(2)+2abx`
`"Then" " " (dy)/(dx)=(d)/(dx)[a^(2)x^(2)+b^(2) + 2abx]`
`=(d)/(dx) (a^(2)x^(2))+(d)/(dx)(b^(2)) +(d)/(dx) (2abx)`
`=a^(2)=(d(x)^(2))/(dx)+ 0+ 2ab(dx)/(dx)`
`=a^(2)xx2x+2ab xx1`
`=2a(ax+b)`


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