If f(x)=(ax+b)n,(x>−ba), then f|(x) is

Write u=ax+b so that f(x)=unThen f|(x)=ddu(un).dudx=nun−1.a=an(ax+b)n−1

If f(x)=(ax+b)n,(x>−ba), then f|(x) is

Write u=ax+b so that f(x)=unThen f|(x)=ddu(un).dudx=nun−1.a=an(ax+b)n−1

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If $f (x) = {(a x + b)}^{n}, (x > \frac{- b}{a})$ , then $f^{|} (x)$ is

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$a n {(a x - b)}^{n - 1}$

$a n {(a x - b)}^{n + 1}$

$a n {(a x + b)}^{n - 1}$

$a n {(a x + b)}^{n + 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} Write u = a x + b so that f (x) = u^{n} \\ Then f^{|} (x) = \frac{d}{d u} (u^{n}) . \frac{d u}{d x} \\ = n u^{n - 1} . a = a n {(a x + b)}^{n - 1} \end{array}$

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