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What the "steps" are depends on what methods you have learned. Part way through a Calculus course you probably would have learned the "power rule", dxndx=nxn1\displaystyle \frac{dx^n}{dx}= nx^{n-1}, and that daf(x)dx=adfdx\displaystyle \frac{daf(x)}{dx}= a\frac{df}{dx}. If you know that then you know that daxndx=naxn1\displaystyle \frac{dax^n}{dx}= nax^{n-1}.


If you do not know those then you will need to use the basic definition of the derivative: dfdx=limh0f(x+h)f(x)h\displaystyle \frac{df}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}.

With f(x)=axn\displaystyle f(x)= ax^n then f(x+h)=a(x+h)n\displaystyle f(x+ h)= a(x+ h)^n. That's why Subhotash Kahn asked "Can you expand: (x + h)^n)?".

You can use the "binomial expansion": (x+h)n=(ni)xihni\displaystyle (x+ h)^n= \sum\begin{pmatrix}n \\ i\end{pmatrix}x^ih^{n-i} where (ni)\displaystyle \begin{pmatrix}n \\ i\end{pmatrix} is the "binomial coefficient" n!i!(ni)\displaystyle \frac{n!}{i!(n-i)}.

That can be expanded as xn+nxn1h+n(n1)2xn2h2+\displaystyle x^n+ nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot. Then f(x+h)f(x)=xn+nxn1h+n(n1)2xn2h2+xn=nxn1h+n(n1)2xn2h2+\displaystyle f(x+h)- f(x)= x^n+ nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot- x^n= nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot.

Notice the "h" in the first term, h2\displaystyle h^2 in the next term, and higher powers of h in succeeding terms. When we divide by h we get nxn1+n(n1)2xn2h2+\displaystyle nx^{n-1}+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot where there is no h in the first term but powers of h in every term after that. Taking the limit as h goes to 0, all terms except the first term goes to 0.

The limit, as h goes to 0, is nxn1\displaystyle nx^{n-1}.
 
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