# A curve has the equation y= (x^2+1)^4 + 2(x^2+1)^3.1. A curve has the equation y= (x^2+1)^4 + 2(x^2+1)^3. Show that dy/dx = 4x(x^2+1)(2x^2+5) and hence show that the curve has just onestationary...

A curve has the equation y= (x^2+1)^4 + 2(x^2+1)^3.

1. A curve has the equation y= (x^2+1)^4 + 2(x^2+1)^3. Show that dy/dx = 4x(x^2+1)(2x^2+5) and hence show that the curve has just one

stationary point. State the coordinates of the stationary point.

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### 1 Answer

You need to prove that `(dy)/(dx) = 4x(x^2+1)(2x^2+5)` , hence, you need to differentiate the given function `y = (x^2+1)^4 + 2(x^2+1)^3` , with respect to x, using the chain rule, such that:

`(dy)/(dx) = 4(x^2 + 1)^3*(x^2 + 1)' + 2*3*(x^2 + 1)^2*(x^2 + 1)'`

`(dy)/(dx) = 4(x^2 + 1)^3*(2x) + 6*(x^2 + 1)^2*(2x)`

`(dy)/(dx) = 8x(x^2 + 1)^3 + 12x(x^2 + 1)^2`

You need to factor out `4x(x^2 + 1)^2` such that:

`(dy)/(dx) = 4x(x^2 + 1)^2(2(x^2 + 1) + 3)`

`(dy)/(dx) = 4x(x^2 + 1)^2(2x^2 + 2 + 3)`

`(dy)/(dx) = 4x(x^2 + 1)^2(2x^2 + 5)`

**Hence, differentiating with respect to x yields** `(dy)/(dx) = 4x(x^2 + 1)^2(2x^2 + 5).`

You need to prove that the function has one stationary point, hence, you need to test if first order derivative has one solution, such that:

`(dy)/(dx) = 0 => 4x(x^2 + 1)^2(2x^2 + 5) = 0`

Since `x^2 + 1 > 0` and `2x^2 + 5 > 0` yields `4x = 0 => x = 0.`

Substituting 0 for x in equation of function yields:

`y = (0 + 1)^4 + 2(0 + 1)^3 = 3`

**Hence, the function has one stationary point yields at `(0,3).` **