Derivative of te^(-t²) and z^(1/2)/e^z using chain rule

[bjon-07]

I am in math 50 (calc one) and i would greatly appreate it if someone could please show me how to solve these problems. I have the anwsers, but i can't figure out how to get them. thank you

In both problems i am need to find the derivative of the function,(we just learned the chain rule, so i am 'supposed to' use that to figure them out. But if you know a another way, great.

problem one

y=te^(-t^2) i normally know how to solve this problems but the t before the e is messing me up. the answer is e^(-t^2)(1-2t^2)

second problem

y=(z^(1/2)/e^z) the answer is e^-z/(2(z^(1/2))-(z^(1/2))e^-Z

thank you again for your help.

 

[KLscilevothma]

Originally posted by bjon-07
In both problems i am need to find the derivative of the function,(we just learned the chain rule, so i am 'supposed to' use that to figure them out. But if you know a another way, great.

problem one

y=te^(-t^2) i normally know how to solve this problems but the t before the e is messing me up. the answer is e^(-t^2)(1-2t^2)

The following two formulae are useful

1. d/dx e^x = e^x

2. d/dx e^g(x)= g'(x) e^g(x) (this is an example of chain rule, if you need further explanation, please tell.)

I'll do the first question and the second one is similar to the first one.

Let
f(t) = te^(-t2)

f'(t) = t d/dt(e^(-t2)) + e^(-t2)d/dt (t)

= t*(-2t)* e^(-t2) (use formula 2 above) + e^(-t2)

= e^-t2[1-2t²]

 

[HallsofIvy]

K L Kam did a good job so I'll just content my self with pointing out, since bjon-07 said specifically that it was the "t before the e" that was giving him trouble: use the PRODUCT rule!
(fg)'= f g'+ f' g

KL Kam used it when he said
"f'(t) = t d/dt(e(-t2)) + e(-t2)d/dt (t)"

 

[bjon-07]

Thank you for your help

Thank you for your help. I was trying to solve the whole thing using only the chain rule ( i did not use the product rule).