InterviewSolution
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InterviewSolution
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Differentiate each of the following w.r.t. x: (i)`(ax+b)^(m)" "(ii)(2x+3)^(5)" "(iii)sqrt(ax^(2)+2bx+c)` |
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Answer» (i) Let `y=(ax+b)^(m)`. Putting `(ax+b)=t`, we get `y=t^(m) and t=(ax+b)` `rArr" "(dy)/(dt)=mt^(m-1)and (dt)/(dx)=a` `rArr" "(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(mt^(m-1)xxa)=mat^(m-1)=ma(ax+b)^(m-1)` `therefore" "(d)/(dx)(ax+b)^(m)=ma(ax+b)^(m-1)`. Let `y=(2x+3)^(5)`. Putting `(2x+3)=t`, we get `y=t^(5)and t=2x+3` `rArr" "(dy)/(dt)=5t^(4)and (dt)/(dx)=2` `rArr" "(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=10t^(4)=10(2x+3)^(4)`. (iii) Let `y=sqrt(ax^(2)+2bx+c).` Putting `(ax^(2)+2bx+c)=t`, we get `rArr" "(dy)/(dx)=(1)/(2)t^(-1//2)=(1)/(2sqrtt)and (dt)/(dx)=(2ax+2b)=2(ax+b)` `rArr" "(dy)/(dx)=((dy)/(dt)xx(dt)/(dx))=(1)/(2sqrtt)xx2(ax+b)` `=((ax+b))/(sqrtt)=((ax+b))/(sqrt(ax^(2)+2bx+c)).` |
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