Thread: Differentiating E

I'm really not understanding how to differentiate things when E to the power of something is present.

For example:

Differentiate:

X^2 times e^-x

Apply Product Rule:

sol'n: 2xe^-x + x^2 ??

The e^-x is confusing me.

Originally Posted by Xavier20

I'm really not understanding how to differentiate things when E to the power of something is present.

For example:

Differentiate:

X^2 times e^-x

Apply Product Rule:

sol'n: 2xe^-x + x^2 ??

The e^-x is confusing me.

let u be a function

d/du e^u = u' e^u

in other words, to differentiate e raised to any power, you multiply e raised to the original power by the derivative of the power.

example:

d/dx e^(x^2) = 2xe^(x^2) ..........2x is the derivative of x^2

now try the question again

X^2 times e^-x

2xe^-x + x^2 times -xe^-x

?

Originally Posted by Xavier20

X^2 times e^-x

2xe^-x + x^2 times -xe^-x

?

no, the derivative of -x is -1 not -x

try one more time

Ohh. Right.

2xe^-x + x^2 times -e^-x

Originally Posted by Xavier20

Ohh. Right.

2xe^-x + x^2 times -e^-x

yeah, but you wouldn't write "times". write it out like a mathematician

2xe^-x + (x^2)(-e^-x)

Originally Posted by Xavier20

2xe^-x + (x^2)(-e^-x)

well that's better, but this is even better

2xe^-x - (x^2)e^-x

or to factorize:

(xe^-x)(2 - x)

try these (if you have time, if you're doing homework otherwise, don't bother):

Find the derivatives of the following:

y = e^(ax^3) .........where a is a constant

y = e^(tsin(2t))

y = sqrt(1 + 2e^(3x))

y = e^(e^x)

y = cos(e^(pi*x))

Find the derivative of the following:

y = e^(ax^3) .........where a is a constant

y' = (3ax^2)e^(ax^3)

y = e^(tsin(2t))

y' = [sin(2t)tcos(2t)]e^(tsin(2t))

y = sqrt(1 + 2e^(3x))

y = (1+2e^(3x))^1/2
y' = [1/2(1+2e^(3x))^-1/2]6e^(3x)

y = e^(e^x)

y' = (e^x)e^(e^x)

y = cos(e^(pi*x))

y' = [-sin(e^(pi*x))]pi*e^(pi*x)

Originally Posted by Xavier20

y = e^(tsin(2t))

y' = [sin(2t)tcos(2t)]e^(tsin(2t))

Good! all were correct except the one above, try to see where you made the mistake

y = e^(tsin(2t))

y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

I forgot the chain rule part of cos(2t).

Originally Posted by Xavier20

y = e^(tsin(2t))

y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

I forgot the chain rule part of cos(2t).

you need both the product and chain rule to find the derivative of tsin(2t)

you have 2 functions in a product, one is t and one is sin(2t) so you need the product rule. but sin(2t) is a composite function, so you need the chain rule for that

Product rule:

d/dx f(x)*g(x) = f ' (x)*g(x) + g'(x)*f(x)

Chain rule:

d/dx f(g(x)) = f ' (g(x))*g'(x)

here goes:

d/dx tsin(2t) = sin(2t) [i took the derivative of t here and multiplied by sin(2t)] + t(2cos(2t)) [i left the t and took the derivative of sin(2t) using the chain rule]

so d/dx tsin(2t) = sin(2t) + 2tcos(2t)

now you can complete the problem

Looks like you got trouble with "e" number

Always remember that



Could be you are messin' it with chain rule.

y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

Yeah. I forgot to add the + sign between the sin(2t) and tcos(2t) and didn't bring the *2 out infront of the tcos(2t) in my chain rule.

Thanks for sorting that out though.

It should be:

y' = [sin(2t) + 2tcos(2t)]e^(tsin(2t))