Deriving the Derivative of y=(x^2+1)^4 + 2(x^2+1)^3

[studentxlol]

Homework Statement

A curve has equation $$y=(x^2+1)^4 + 2(x^2+1)^3$$. Show that $$\frac{dy}{dx}=4x(x^2+1)^2(2x^2+5)$$.

Homework Equations

$$\frac{dy}{dx}=(\frac{dy}{du}\times \frac{du}{dx})+(\frac{dy}{dv} \times \frac{dv}{dx})$$

The Attempt at a Solution

$$y=(x^2+1)^4 + 2(x^2+1)^3$$

$$let u = (x^2+1)^4 and v=x^2+1 so that y=u^4+v^3$$

$$\frac{dy}{du}=4u^3=4(x^2+1)^3 and \frac{du}{dx}=2x$$

$$\frac{dy}{dv}=3v^2=3(x^2+1)^2 and \frac{dv}{dx}=2x$$

$$\therefore \frac{dy}{dx}=(\frac{dy}{du}\times \frac{du}{dx})+(\frac{dy}{dv} \times \frac{dv}{dx})=8x(x^2+1)^3+6x(x^2+1)^2$$

Why am I not getting the answer $$4x(x^2+1)^2(2x^2+5)$$?

 

[second]

Homework Statement

A curve has equation $$y=(x^2+1)^4 + 2(x^2+1)^3$$. Show that $$\frac{dy}{dx}=4x(x^2+1)^2(2x^2+5)$$.

Homework Equations

$$\frac{dy}{dx}=(\frac{dy}{du}\times \frac{du}{dx})+(\frac{dy}{dv} \times \frac{dv}{dx})$$

The Attempt at a Solution

$$y=(x^2+1)^4 + 2(x^2+1)^3$$

$$let u = (x^2+1)^4 and v=x^2+1 so that y=u^4+v^3$$

$$\frac{dy}{du}=4u^3=4(x^2+1)^3 and \frac{du}{dx}=2x$$

$$\frac{dy}{dv}=3v^2=3(x^2+1)^2 and \frac{dv}{dx}=2x$$

$$\therefore \frac{dy}{dx}=(\frac{dy}{du}\times \frac{du}{dx})+(\frac{dy}{dv} \times \frac{dv}{dx})=8x(x^2+1)^3+6x(x^2+1)^2$$

Why am I not getting the answer $$4x(x^2+1)^2(2x^2+5)$$?

WELL YOU HAVE TO FACTOR IT. 8x(x^2+1)^3+12x(x^2+1)^2
IS THE SAME AS 4X [2(X^2+1)^3 +3(X^2+1)^2]...KEEP DOING IT..AT THE END YOU WILL HAVE THE ANSWER

BY THE WAY IT SHOULD BE 12 X NOT 6X

 

[HallsofIvy]

I would start by writing
$$y=(x^2+1)^4+2(x^2+1)^3= (x^2+ 1)(x^2+ 1)^3+ 2(x^2+ 1)^3= (x^2+ 1)^3(x^2+ 3)$$.

Now,
$$dy/dx= 3(x^2+ 1)^2(2x)(x^2+ 3)+ (x^2+ 1)^3(2x)$$