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.
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)
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
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))