Question

what is the first derivative of sin^2(3t^2) + 3e^(-t) - 5t^2 cos (2t^3)??

Answer 1

yes, of course

Answer 2

well the first one can be written like this\[(\sin(3t^2))^2\] and then you can use the power rule and chain rule. the second is just the chain rule and the third is the product rule with the chain rule.

Answer 3

\[2(\sin(3t^2)(\cos(3t^2))(6t)\]

Answer 4

so for \[3e ^{-t} = -3e ^{-t}\]
right?

Answer 5

Given sin^2(3t^2) + 3e^(-t) - 5t^2 cos(2t^3)
differentiate with respect to
=>d/dt sin²(3t²)+d/dt 3e^(-t)-d/dt5t^2 cos (2t^3)
=> -10 t cos(2t^3)|+6t sin (6t²)+30t^4 sin (2t^3)-3e^-t

Learn more about Concept of Standard Deviation at http://goo.gl/rGnw84

Answer 6

aishh, many thankz @JMark , i understand now :D