Isaac Boakye
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What are the steps in differentiating y=ax^n
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- Thread starter Isaac Boakye
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What are the steps in differentiating y=ax^n
Can you expand:
(x + h)^n
HallsofIvy
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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", \(\displaystyle \frac{dx^n}{dx}= nx^{n-1}\), and that \(\displaystyle \frac{daf(x)}{dx}= a\frac{df}{dx}\). If you know that then you know that \(\displaystyle \frac{dax^n}{dx}= nax^{n-1}\).
If you do not know those then you will need to use the basic definition of the derivative: \(\displaystyle \frac{df}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}\).
With \(\displaystyle f(x)= ax^n\) then \(\displaystyle f(x+ h)= a(x+ h)^n\). That's why Subhotash Kahn asked "Can you expand: (x + h)^n)?".
You can use the "binomial expansion": \(\displaystyle (x+ h)^n= \sum\begin{pmatrix}n \\ i\end{pmatrix}x^ih^{n-i}\) where \(\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}\) is the "binomial coefficient" \(\displaystyle \frac{n!}{i!(n-i)}\).
That can be expanded as \(\displaystyle x^n+ nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot\). Then \(\displaystyle f(x+h)- f(x)= x^n+ nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot- x^n= nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot\).
Notice the "h" in the first term, \(\displaystyle h^2\) in the next term, and higher powers of h in succeeding terms. When we divide by h we get \(\displaystyle nx^{n-1}+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot\) 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 \(\displaystyle nx^{n-1}\).
If you do not know those then you will need to use the basic definition of the derivative: \(\displaystyle \frac{df}{dx}= \lim_{h\to 0}\frac{f(x+ h)- f(x)}{h}\).
With \(\displaystyle f(x)= ax^n\) then \(\displaystyle f(x+ h)= a(x+ h)^n\). That's why Subhotash Kahn asked "Can you expand: (x + h)^n)?".
You can use the "binomial expansion": \(\displaystyle (x+ h)^n= \sum\begin{pmatrix}n \\ i\end{pmatrix}x^ih^{n-i}\) where \(\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}\) is the "binomial coefficient" \(\displaystyle \frac{n!}{i!(n-i)}\).
That can be expanded as \(\displaystyle x^n+ nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot\). Then \(\displaystyle f(x+h)- f(x)= x^n+ nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot- x^n= nx^{n-1}h+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot\).
Notice the "h" in the first term, \(\displaystyle h^2\) in the next term, and higher powers of h in succeeding terms. When we divide by h we get \(\displaystyle nx^{n-1}+ \frac{n(n-1)}{2}x^{n-2}h^2+ \cdot\cdot\cdot\) 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 \(\displaystyle nx^{n-1}\).