Question

Anyone who knows derivatives?I need help because i'm having a hard time understanding derivatives having trigo functions and derivative of square root.

Answer 1

how to solve this \[\sqrt{x+1}\]

Answer 2

Use the chain rule

Answer 3

u r wrong........................ zhanga

Answer 4

Crap I am.

Answer 5

but anyway, use chain rule.

Answer 6

zhanga was right about applying chain rule but wrong with the answer hahahaha.

let u = x + 1
du/dx = 1

dy/dx = dy/du * du/dx

dy/dx(sqrt(u)) = 1 / (2(sqrt(x + 1))) * 1

therefore the derivative of sqrt(x+1) is 1 / (2sqrt(x + 1))

Answer 7

\[y=\sqrt{x+1}\]Let u = x+1
Therefore, du/dx = 1
and dy/du = 1/2u^-1/2
To find dy/dx you have to multiply dy/du and du/dx,
Therefore,
dy/dx = 1/2(x+1)^-1/2

Answer 8

but how to solve it using the 4 step rule?

Answer 9

come to twiddla

Answer 10

http://www.twiddla.com/584540

Answer 11

easy way:
sqrt(x+1)
(x+1)^(1/2)
The step above is changing from square root form to exponent form. Something to the one half power means "square root it".

So now we use chain rule:
1/2(x+1)^(-1/2) is the first step. All we did was power rule: drop the power in front, and subtract 1/2 by "one", which would be 2/2 in fraction form, leaving us with negative one half.

But we're not done yet. We have to find the derivative of the inside junk (x+1). That would be 1+0 (x turns into 1, and the +1 just gets turned into 0.

So:

1/2 ((x+1) ^(1/2))*1

or...

\[\frac{1}{2\sqrt{x+1}}\]

Answer 12

*So: 1/2 ((x+1) ^(-1/2))*1
FOrgot to put my negative sign in my original "so: blah blah" thing. The final answer is still the same, though.

Answer 13

hey sheena are you done with diff. calculus?