Difference between revisions of "Test"

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\frac{\mathrm d}{\mathrm d x} \big( I(x) \big)
 
\frac{\mathrm d}{\mathrm d x} \big( I(x) \big)
 +
 +
 +
\frac{\mathrm d}{\mathrm d x} \big( I(x) \big)=-\alphaci(x)
 
</math>
 
</math>

Revision as of 21:59, 3 February 2017

[math] x^n + y^n = z^n k_{n+1} = n^2 + k_n^2 - k_{n-1} \frac{\mathrm d}{\mathrm d x} \big( I(x) \big) \frac{\mathrm d}{\mathrm d x} \big( I(x) \big)=-\alphaci(x) [/math]