Ask your own question, for FREE!
Mathematics 25 Online
OpenStudy (anonymous):

Derivative of x^r * e^sx

zepdrix (zepdrix):

\[\Large\bf\sf x^r\cdot e^{sx}\]We need to take the derivative of this with respect to x?

OpenStudy (anonymous):

yes. The r throws me off

zepdrix (zepdrix):

TEAM JACOBBBBBB!!!!

zepdrix (zepdrix):

I'm jkjk jk lol.. ive never actually seen the movies.. just read your profile XD trying to stir up trouble.

zepdrix (zepdrix):

ok ok anyway, the problem here.. let's see

zepdrix (zepdrix):

\[\Large\bf\sf \left(x^r\cdot e^{sx}\right)'\quad=\quad \color{royalblue}{\left(x^r\right)'}e^{sx}+x^r\color{royalblue}{\left(e^{sx}\right)'}\]We have the product of two functions of x. So here is our product rule setup. We need to take the derivative of the blue parts. The first one is giving you trouble?

OpenStudy (anonymous):

yes, the first part

OpenStudy (anonymous):

Would it be 0 because of r or no?

zepdrix (zepdrix):

Just power rule silly! :O

zepdrix (zepdrix):

\[\Large\bf\sf (x^5)'\quad=\quad 5x^{5-1}\]

zepdrix (zepdrix):

\[\Large\bf\sf (x^r)'\quad=\quad ?\]

OpenStudy (anonymous):

rx^r-1?

zepdrix (zepdrix):

good good good.

zepdrix (zepdrix):

\[\Large\bf\sf \left(x^r\cdot e^{sx}\right)'\quad=\quad \color{orangered}{\left(rx^{r-1}\right)}e^{sx}+x^r\color{royalblue}{\left(e^{sx}\right)'}\]

zepdrix (zepdrix):

Understand how to do the other blue part?

OpenStudy (anonymous):

ohh! okay! the other is se^sx

zepdrix (zepdrix):

That almost looks like a bad word!! :O But yes good job! :)

OpenStudy (anonymous):

haha! Thank You so much!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!