Question

derivative of 2x(x^2+1)^4dx

Answer 1

i think it is antiderivative or integral .\[I=\int\limits 2x \left( x^2+1 \right)^4dx\]
put \[x^2+1=t,2x~dx=dt\]
\[I=\int\limits t^4 dt=\frac{ t^5 }{ 5 }+c\]
replace the value of t

Answer 2

Face palm.

Answer 3

Just in case the \(dx\) is placed accidentally in there, and this is indeed a problem that requires to differentiate the expression, I will give you a few examples.
\(\text{___________________________________________________}\)
But, before I give examples, note please the following:

Suppose you had a function
\(\color{black}{\displaystyle y(x)=f(x)\cdot {\rm h}[g(x)]}\)

Then, using the "Product Rule", and the "Chain Rule", you will have:
\(\color{black}{\displaystyle y'(x)=f'(x)\cdot {\rm h}[g(x)]+f(x)\cdot {\rm h}'[g(x)]\cdot g'(x)}\)
\(\text{___________________________________________________}\)

Example 1:
\(\color{black}{\displaystyle f(x)=ax(x^n+m)^{\beta};~~~\beta\ne-1}\)

\(\color{black}{\displaystyle f'(x)=a(x^n+m)^{\beta}+a x\cdot \beta(x^n+m)^{\beta-1}(nx^{n-1})}\)

\(\color{black}{\displaystyle f'(x)=a(x^n+m)^{\beta}+a\beta nx^n (x^n+m)^{\beta-1}}\)
\(\text{___________________________________________________}\)

Example 2:
\(\color{black}{\displaystyle f(x)=2x(e^{3x}+x^6)^{4}}\)

\(\color{black}{\displaystyle f'(x)=2(e^{3x}+x^6)^4+2x\left[4(e^{3x}+x^6)^{3}\cdot (3e^{3x}+6x^5)\right]}\)

\(\color{black}{\displaystyle f'(x)=2(e^{3x}+x^6)^4+24x(e^{3x}+x^6)^{3} (e^{3x}+2x^5)}\)

Answer 4

if there is no dx then certainly it is a problem of derivatives.