How do you use the chain rule to differentiate #y=(x^2+3)^4#?

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Answer 1 · 2018-03-30T03:32:09

#=>y' = 8x(x^2+3)^3 #

#(d)/(dx) f(g(x)) = (df(g(x)))/(dg(x)) * (dg(x))/(dx)#

We have:

So we first compute #(df(g(x)))/(dg(x))#.

#(d(x^2+3)^4)/(d(x^2+3)) = color(blue)(4(x^2+3)^3)#

Now we compute #(dg(x))/(dx)#.

#(d(x^2+3))/(dx) = color(blue)(2x)#

We multiply these two results to get the final solution:

#=>y' = 8x(x^2+3)^3 #

Answer 2 · 2018-03-30T03:52:00

#8x(x^2+3)^3#

The chain rule states that,

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

Let #u=x^2+3#, then #(du)/dx=2x#.

We have: #y=u^4#, and therefore #dy/(du)=4u^3#.

Multiply these results together to get,

#dy/dx=4u^3*2x#

#=8xu^3#

Now, we undo the substitution #u=x^2+3#, and we get:

#=8x(x^2+3)^3#