Extending Differentiation (Question)

Question

A curve has equation y=(x²+1)⁴+2(x²+1)³. Show that dy/dx=4x(x²+1)² (2x²+5) and hence show that the curve has one stationary (critical) point. State the coordinates of the stationary point and, by considering the gradient of the curve on either side of the stationary point, determine its nature. 

Answer 1

Take the derivative of y using the chain-rule.

 

dy/dx = 4(x² + 1)³(2x) + 6(x² + 1)²(2x)

 

dy/dx = 8x(x² + 1)³ + 12x(x² + 1)²

 

 

Factor out 4x(x² + 1)².

 

dy/dx = 4x(x² + 1)²[2(x² + 1) + 3]

 

dy/dx = 4x(x² + 1)²(2x² + 5)

 

 

Now set dy/dx=0 to find the critical point.

 

4x(x² + 1)²(2x² + 5) = 0

 

 

The last two factors give you a complex solution when set to zero.  Therefore,

 

4x = 0

 

   x = 0     ----> critical value

 

 

Evaluate  f'(-1) and f'(1).   These are the gradients on either side of the critical value.

 

If these derivatives have different signs, then f(0) is a critical point.

 

Answer 2

y = f(x) = (x²+1)⁴ + 2(x²+1)³

 

           = (x²+1)³[(x²+1)+2]

 

            = (x²+1)³(x²+3)

 

By the product rule, y' = 3(x²+1)²(2x)(x²+3) + 2x(x²+1)³

 

                               = 2x(x²+1)²[3(x²+3)+x²+1]

 

                               = 2x(x²+1)²(4x²+10)

 

                               = 4x(x²+1)²(2x²+5)

 

x²+1 and 2x²+5 are nonzero if x is real

 

So, if y'=0 then 4x = 0.  Therefore, the only critical number is x = 0.

 

Critical point:  (0, f(0)) = (0, 3)

 

If x<0, then y'<0 

 

If x>0, then y'>0

 

So, the point (0,3) is a relative minimum.