Question

Производные с подробным решением (x^6-7)' (4x+1/x)' ((2x+3)*(4x^2-3x+7))' (x^2-2x/x+4x^3)

Answer 1

$(x^6-7)'=6x^5\\\\\\(4x+\frac{1}{x})'=4+(x^{-1})'=4-x^{-2}=4-\frac{1}{x^2}\\\\\\((2x+3)(4x^2-3x+7))'=2(4x^2-3x+7)+(2x+3)(8x-3)\\\\\\(\frac{x^2-2x}{x+4x^3})'=\frac{(2x-2)(x+4x^3)-(x^2-2x)(1+12x^2)}{(x+4x^3)^2}=\\\\=\frac{2x^2+8x^4-2x-8x^3-x^2+2x-12x^4+24x^3}{(x\cdot (1+4x^2))^2}=\frac{-4x^4+16x^3+x^2}{x^2(1+4x^2)^2}=\frac{-4x^2+16x+1}{(1+4x^2)^2}$