Differentiate (ax+b)raised to the power n by abinitio method?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
Use the limit definition and make use of the binomial expansion
This perhaps belongs more in Calculus than in Physics?
Problem: Deduce #d/dx[(ax+b)^n]#
I assume that by "ab initio" you mean to derive it in terms of infinitesimal limits?
#(df(x))/dx=lim_(h->0)(f(x+h)-f(x))/((x+h)-x)=lim_(h->0)(f(x+h)-f(x))/h#
If #f(x)=(ax+b)^n#, then
#(df)/dx=lim_(h->0)((a(x+h)+b)^n-(ax+b)^n)/h#
#=lim_(h->0)((ah+(ax+b))^n-(ax+b)^n)/h#
If we expand the first bracket via the binomial theorem, we obtain
#(ah+(ax+b))^n=(ah)^n+n(ah)^(n-1)(ax+b)+1/2n(n-1)(ah)^(n-2)(ax+b)^2+...+1/2n(n-1)(ah)^2(ax+b)^(n-2)+n(ah)(ax+b)^(n-1)+(ax+b)^n#
Note that the final term of the expansion cancels with the #-(ax+b)^n# in the numerator of our limit expression, leaving only terms containing a factor of #h#.
#(df)/dx=lim_(h->0)((ah)^n+...+1/2n(n-1)(ah)^2(ax+b)^(n-2)+n(ah)(ax+b)^(n-1)
)/h#
#=lim_(h->0)(a^nh^(n-1))+...+1/2n(n-1)(a^2h)(ax+b)^(n-2)+na(ax+b)^(n-1)
#
As #h->0#, then every term disappears apart from the last one, leaving us with
#(df)/dx=na(ax+b)^(n-1)#
