Question

Помогите с решением,заранее спасибо!

Answer 1

$y=\frac{3x^9-2x^5}{13-2x^2}\\\\y'=\frac{(27x^8-10x)(13-2x^2)-(3x^9-2x^5)(-4x)}{(13-2x^2)^2}=\\\\=\frac{351x^8-54x^{10}-130x+20x^3+12x^{10}-8x^6}{(13-2x^2)^2}=\frac{351x^8-42x^{10}-130x+20x^3-8x^6}{(13-2x^2)^2}\\\\\\\star \; \; (x^{k})'=k\cdot x^{k-1}\; ,\; \; (\frac{u}{v})'= \frac{u'v-uv'}{v^2}\; \; \star$

Answer 2

y'= [ (3*9x^8 - 2*5x^4)*(13-2x^2) -(3x^9-2x^5)*(-4x) ] / (13-2x^2)^2