find derivative of y=(x^2+1)(x^3+1) by using Product Rule?

... d(ab) = a d(b) + b d(a)

f(x) = (x^2+1) (x^3+1)

d[ f(x) ] = (x^2+1) d [ (x^3+1) ] + (x^3+1) d [ (x^2+1) ]

.......... = (x^2+1) [3x^2] d [x] + (x^3+1) [2x] d [x]

df/dx = (x^2+1) [3x^2] + (x^3+1) [2x]

....... = 3x^4 + 3x^2 + 2x^4 + 2x

....... = 5x^4 + 3x^2 + 2x

to get that answer what you need to do is first

(x^2+1)*dx(x^3+1)+(x^3+1)*dx(x^2+1)

(x^2+1)*3x^2+(x^3+1)*2x

so it will be

3x^4+3x^2+2x^4+2x

add the like terms

5x^4+3x^2+2x

and if you want you can factor one x to make it a bit more simple

First, find your derivatives.

In operator notation:

Dx [x^2 + 1 ] = 2x.

Dx [x^3 + 1 ] = 3x^2.

Now we have f.Dx[g] + g.Dx[f] =

(3x^2)(x^2 + 1) + (x^3 + 1)(2x).

= 3x^4 + 3x^2 + 2x^4 + 2x.

= 5x^4 + 3x^2 + 2x.

y' = (x² + 1)(3x²) + (x³ + 1)(2x)

y' = 3x^4 + 3x² + 2x^4 + 2x

y' = 5x^4 + 3x² + 2x

Use product rule. there is likewise some chain rule in looking the by-manufactured from (6-4x)^12 it is going to be the 1st circumstances the by-manufactured from the 2nd plus the 2nd circumstances the by-manufactured from the 1st: (x^8)*(a million/2)(6-4x)^(-a million/2)*(-4)+(6-4x)^a million... *(8x^7) i don't understand your alternatives although...