Investigation

## Solution for $x^n$

How can we generalise to find the gradient for any tangent line to a function $y=x^n$, where $n$ is a positive integer?

We could adapt what we did for the quadratic and cubic curves in order to generalise: how can we find the gradient of the line through the points with coordinates $(x,x^n)$ and $(x+h,(x+h)^n)$? Where $h$ is a very very small positive number.

We can then calculate the gradient of the line going through the points $(x,x^n)$ and $(x+h,(x+h)^n)$,

$\frac{y_2-y_1}{x_2-x_1}=\frac{(x+h)^n-x^n}{x+h-x}=\frac{(x+h)^n-x^n}{h}.$

We would like to expand and simplify this expression, but $(x+h)^n$ is not so simple to expand when we don’t have a specific value for $n$.

However, $x+h$ is a binomial. We have a way to expand $(x+h)^n$, The Binomial Theorem, for $n$ a positive integer:

$(x+h)^n={\binom n 0}x^nh^0+{\binom n 1}x^{n-1}h+{\binom n 2}x^{n-1}h^2+ \dots + {\binom n {n-1}}xh^{n-1}+{\binom n n}x^{0}h^n.$

The binomial coefficients ${\dbinom n k}$ are given by $\dfrac{n!}{k!(n-k)!}$.

We can now think about how to expand,

$\frac{(x+h)^n-x^n}{h}.$

The numerator is,

${\binom n 0}x^nh^0+{\binom n 1}x^{n-1}h+{\binom n 2}x^{n-1}h^2+ \dots + {\binom n {n-1}}xh^{n-1}+{\binom n n}x^{0}h^n-x^n$

which simplifies to,

${\binom n 1}x^{n-1}h+{\binom n 2}x^{n-1}h^2+ \dots + {\binom n {n-1}}xh^{n-1}+{\binom n n}x^{0}h^n.$

The denominator simplifies to $h$.

So, the gradient of the line through the points $(x,x^n)$ and $(x+h,(x+h)^n)$ is

${\binom n 1}x^{n-1}+{\binom n 2}x^{n-1}h+ \dots + {\binom n {n-1}}xh^{n-2}+{\binom n n}h^{n-1}.$

As we move the point $(x+h,(x+h)^n)$ closer and closer to $(x,x^n)$, we are making $h$ smaller and smaller, finally making $h$ zero. This gives the gradient of the line that touches the curve $y=x^n$ at the point $(x,x^n)$ as ${\binom n 1}x^{n-1}$, for $n$ a positive integer.

The binomial coefficient $\dbinom{n}{1}$ is given by $\dfrac{n!}{1!(n-1)!}=\dfrac{n\times (n-1) \times \dots \times 1}{(n-1)\times (n-2) \times \dots \times 1}=n$.

This gives the gradient of the line that touches the curve $y=x^n$ at the point $(x,x^n)$ as $nx^{n-1}$, for $n$ a positive integer.