Proving Differentiation rule using Induction

[oshea.emma]

Hi Everyone!!
Any advice on this question?

Prove by induction that d/dx(x^n)equals nx^n-1 where n greater equal to 1 and n is an element of the natural numbers

What i have so far is
For n = 1, d(x)/dx = 1x^(1-1)

Assuming true @ n = k, it works too: d(x^k) = kx^k-1.

Proving true for n=k+1 is where i'm stuck at? Maybe someones done one of these questions before?

:?

 

[pka]

Given that the following is true,
\(\displaystyle \L
\frac{{d\left( {x^K } \right)}}{{dx}} = Kx^{K - 1}\)
then we note that \(\displaystyle \L
x^{K + 1} = \left( x \right)\left( {x^K } \right)\).

Using the product rule:
\(\displaystyle \L
\begin{array}{rcl}
\frac{{d\left( {x^{K + 1} } \right)}}{{dx}} & = & \frac{{d\left[ {(x)\left( {x^K } \right)} \right]}}{{dx}} \\
& = & \frac{{d\left[ {(x)} \right]}}{{dx}}\left( {x^K } \right) + (x)\frac{{d\left[ {\left( {x^K } \right)} \right]}}{{dx}} \\
& = & (1)\left( {x^K } \right) + (x)\left[ {Kx^{K - 1} } \right] \\
& = & \left( {x^K } \right) + \left[ {Kx^K } \right] \\
& = & (K + 1)x^K \\
& = & (K + 1)x^{(K + 1) - 1} \\
\end{array}\)

 

[oshea.emma]

Excellent answer!
Very elegant! Thanks very very much!!
It makes a lot of sense when it's done out like that!