derivative problems: y = e^(-x^2), g(t) = e^(-3/t^2)

[SMAlvarez]

I am not sure what to do with the these problems, although they may seem simple:

1) y = e^(-x^2)

2) g(t) = e^(-3/t^2)

Thanks for any and all help.

 

[tkhunny]

Re: Simple derivative problems

I am not sure what to do with the these problems, although they may seem simple:

1) y = e^(-x^2)

2) g(t) = e^(-3/t^2)

Thanks for any and all help.

So, it's Chain Rule day?

If f(x) = g(h(x)), then f'(x) = g'(h(x))h'(x)

You have y = e^(-x^2)

Just identify y = f(x), g(x) = e^x, and h(x) = -x^2

Then, f'(x) = y' = g'(h(x))h'(x) = [e^(-x^2)](-2x)

you do the next one.

 

[SMAlvarez]

ok so g(x) =e^x and h(x) = -3/t^2

So the answer is [e^(-3/t^2)](6x/t^4)

Thats if I did the quotient rule properly on -3/t^2

 

[SMAlvarez]

Did I get it right?

 

[stapel]

ok so g(x) =e^x and h(x) = -3/t^2

Where are the x's coming from? :shock:

So the answer is [e^(-3/t^2)](6x/t^4)

Sorry; no.

Thats if I did the quotient rule properly on -3/t^2

What were your steps? (And why not use the simpler Power Rule, with -3t^-2?) :idea:

Please be complete. Thank you!

Eliz.

 

[SMAlvarez]

I think I get it now, thanks

 

[tkhunny]

And why not use the simpler Power Rule, with -3t^-2?)

That WILL keep you from getting all the way to \(\displaystyle t^{4}\) when you shoud have ended up at \(\displaystyle t^{3}\).