How to differentiate?

Question

#y=(x^4+2x-22)^2#

1304 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-21T13:12:37

Use chain rule to give #dy/dx = 2(x^4 + 2x -22)*(4x^3 + 2)#

#u=x^4 + 2x - 22#
#y= u^2#
#dy = 2u du #
#du = 4x^3 + 2 dx#

#dy/dx = 2(x^4 + 2x -22)*(4x^3 + 2)#

Answer 2 · 2018-03-21T13:17:26

#2(x^4+2x-22)(4x^3+2)#

We use chain rule, which states that,

#(df)/dx=(df)/(du)*(du)/dx#

Let #u=x^4+2x-22#, then #(du)/dx=4x^3+2#. Also, then we have #f=u^2, (df)/(du)=2u#.

Combining, we get:

#(df)/dx=2u*(4x^3+2)#

Now, we need to undo the substitution, and we get

#=2(x^4+2x-22)(4x^3+2)#