Can someone help me with math?
I've just learned about the Chain Rule and I've been finding derivatives without too much of a problem, but I've come to a bind here and I don't understand what I'm doing wrong.
"Differentiate y = sq.rt(x) + 1/4[sin(2x)^2]"
The equation is a sum, so you can just take the derivative of both components of the sum separately. The derivative of sq.rt(x) is 1/2sq.rt(x). I have to use the Chain Rule for the last component.
I like to work out and then go in, so this is what I did:
y' = 1/2*sq.rt(2) + 1/4(cos(2x)^2)) * d/dx (2x)^2
I took the derivative of sine since it's the outermost function.
The derivative of (2x)^2 is 4x.
This gives me:
y' = 1/2*sq.rt(2) + 1/4(cos(2x)^2)) * 4x
y' = 1/2*sq.rt(2) + x(cos(2x)^2)
But according to my book, that's wrong. My book says it's y' = 1/2*sq.rt(2) + 2x(cos(2x)^2)).
What am I doing wrong?
(Just to be clear, it's (2x) that's being squared, it's not sin^2(2x).)
3 Comments • Newest first
Ah, yeah, got it.
Maybe I should just take a break for today
Thanks.
@SunsetDews:
first change the equation to y = x^(1/2) + 1/4 sin(4x^2)
dy/dx = 1/2x^(-1/2) + [1/4cos(4x^2) * 8x]
= 1/2x^(-1/2) + 2xcos(4x^2)
reduces to 1/(2*sqrt(x)) + 2xcos(2x)^2
edit: because (2x)^2 = 2x*2x = 4x^2
[quote=tauniversa]check on the derivative of (2x)^2, also 4x^2[/quote]
Man, this is what I get for rushing...
Thanks.
EDIT: Wait, what?
What are you talking about?