Question

derivative of sqrt(4t^2+4t^4+1)

Answer 1

\[ 4\,{\frac { \left( 1+2\,{t}^{2} \right) t}{\sqrt { \left( 1+2\,{t}^{2}
\right) ^{2}}}} \]

Answer 2

d/dt(sqrt(4 t^4+4 t^2+1))
| Use the chain rule, d/dt(sqrt(4 t^4+4 t^2+1)) = ( dsqrt(u))/( du) ( du)/( dt), where u = 4 t^4+4 t^2+1 and ( dsqrt(u))/( du) = 1/(2 sqrt(u)):
= | (d/dt(4 t^4+4 t^2+1))/(2 sqrt(4 t^4+4 t^2+1))
| Differentiate the sum term by term and factor out constants:
= | (4 (d/dt(t^4))+4 (d/dt(t^2))+d/dt(1))/(2 sqrt(4 t^4+4 t^2+1))
| The derivative of 1 is zero:
= | (4 (d/dt(t^4))+4 (d/dt(t^2))+0)/(2 sqrt(4 t^4+4 t^2+1))
| The derivative of t^2 is 2 t:
= | (4 (d/dt(t^4))+4 (2 t))/(2 sqrt(4 t^4+4 t^2+1))
| The derivative of t^4 is 4 t^3:
= | (4 (4 t^3)+8 t)/(2 sqrt(4 t^4+4 t^2+1))