How do you find the derivative of #u=(x^2+3x+1)^4#?

I'm going to go with the second option, because I'm not a masochist (jokes!)

When you take a derivative of a set of terms in parentheses, you can do a substitution, treating the parentheses as a separate single function. We'll treat the in-parentheses terms like this:

#(x^2+3x+1)=g(x)#

so our function now looks like this:

#u=g(x)^4#

Now, the derivative is not simply #(du)/(dx)=4g(x)^3#, but it is instead #(du)/(dx)=4g(x)^3*(dg)/(dx)#. We must calculate #(dg)/(dx)# before arriving at our solution.

if #g(x)=x^2+3x+1# then:

#(dg)/(dx)=2x+3#

Re-inserting our substitutions, we now arrive at our answer:

#(du)/(dx)=4(x^2+3x+1)^3*(2x+3)#

Use the chain and power rule. Find the derivative of #1(....)^4#(use the power rule. Then find the derivative of #(x^2+3x+1)#(use the power rule). then combine the two. #4(x^2+3x+1)^3(2x+3) #

#"differentiate using the "color(blue)"chain rule"#

#"Given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#u=(x^2+3x+1)^4#

#rArr(du)/dx=4(x^2+3x+1)^3xxd/dx(x^2+3x+1)#

#color(white)(rArr(du)/dx)=4(2x+3)(x^2+3x+1)^3#