Question

nth derivative of y=(ax+b)^m​

Answer 1

Answer:

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Answer 2

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let there be a function

y=(ax+b)^m,

then the first derivative of function “y” with respect to “x” is given by

y1=m(ax+b)^(m-1).a , by chain rule . . .eqn[1]

again differentiating eqn[1] with respect to “x” we get second derivative ,

y2=am(m-1)(ax+b)^(m-2).a =a^2.m(m-1)(ax+b)^(m-2) . . .eqn[2]

similarly differentiating eqn[2] with respect to “x” we get third derivative of function “y” that is given by,

y3=a^2.m(m-1)(m-2)(ax+b)^(m-3)=a^3.m(m-1)(m-2)(ax+b)^(m-3)

for nth derivative

yn=a^(n-1).m(m-1)(m-2). . .{m-(n-3)}{m-(n-2)}{m-(n-1)}(ax+b)^(m-n).a

=a^(n-1).am(m-1)(m-2). . . .{m-(n-1)}(ax+b)^(m-n)

=a^n.m(m-1)(m-2). . .(m-n+1)(ax+b)^(m-n)

hence the nth derivative is

yn=a^n.m(m-1)(m-2). . .(m-n+1)(ax+b)^(m-n)