whats the derivative of y= x^2/(1-x^3)?

Question

also whats the derivative of y=3x+xtanx

Answer 1

1.) Use quotient rule.
dy/dx = [(1 - x^3)(2x) - (x^2)(-3x^2)]/(1 - x^3)^2
dy/dx = [2x +x^4]/(1 + x^3)^2

2.) dy/dx = 3 + tanx + x(sec^2(x))

Answer 2

You have 2 parts, where you derivate the upper part and multiplicate with the lower part. Then you substract the upper part with the derivative of the lower part. These two operations goes in the upper part of the solution. Finally, al this is divided by the lower part ^2
y(x) = u/v
y'(x) = (u'v -uv')/v^2
y'(x) = [2x(1-x^3) - x^2(-3x^2)] / (1-x^3)^2
y'(x) = x(2 + x^3)/(1-x^3)^2

Answer 3

y = x^2 / (1 - x^3)

Use the quotient rule.

dy/dx = [ (2x)(1 - x^3) - (x^2)(-3x^2) ] / (1 - x^3)^2

dy/dx = [ 2x - 2x^4 + 3x^4 ] / (1 - x^3)^2
dy/dx = [ 2x - x^4 ] / (1 - x^3)^2
dy/dx = x(2 - x^3) / (1 - x^3)^2

2) y = 3x + x tan(x)

Use the product rule on the second term.

dy/dx = 3 + (1)tan(x) + x(sec^2(x))
dy/dx = 3 + tan(x) + x sec^2(x)

Answer 4

y'={2x(1-x^3)-(-3x^2. x^2)}/(1-x^3)^2


y'=3+1tanx+(1+tan^2 x ) x

Answer 5

y' = [(1-x^3)(2x)-x^2(-3x^2)]/(1-x^3)^2

y' = (2x+x^4)/(1-2x^3+x^6)

use quotient rule