Question

Find the derivative:
a)square root of x - 3 square root of x
b)3x/(x^2+4)

Answer 1

\[a) \sqrt{x-3}\sqrt{x}\]
\[b)\frac{ 3x }{ x^{2}+4 }\]
Are these your questions?

Answer 2

a) square root of x subtract 3 square root of x
b is the questions

Answer 3

so (a) will be )\[\sqrt{x}-3\sqrt{x}\]

Answer 4

a)\[\sqrt{x}-3\sqrt{x}=-2\sqrt{x}=-2x^{\frac{ 1 }{ 2 }}\]
then find derivative easily

Answer 5

b) use (Quotient Rule) :

Answer 6

@cutepochacco do you understand ?

Answer 7

Hey, cute, there's some confusion here about the identity of the function whose derivative you need to find. Have you ever tried using the equation editor (see the icon below)? There's a learning curve, but once you know how to use that editor, it's often much easier to be clear in defining functions.

Answer 8

\[f(x)=\sqrt{x}-3\sqrt{x} \]
\[f(x)=\sqrt{2x}\]
\[f(x)=(x^{2}+2)(2x ^{3}-5x ^{2}+4x)\]

Answer 9

Cool, cool, cool! Thank you.
#1: rewrite this function as
\[f(x)=1\sqrt{x}-3\sqrt{x}=-2\sqrt{x}=-2x ^{(1/2)}.\]

Now apply the power and constant coefficient rules:
\[f'(x)=-2*\frac{ 1 }{ 2 }x ^{(\frac{ 1 }{ 2 }-1)}=-x ^{(-\frac{ 1 }{ 2 })}.\]

Answer 10

Let's discuss this. Is there any step involved here that you'd like to discuss?

Answer 11

Note that much of the work involved in solving this problem was in re-writing the original expression.

Answer 12

I hope this helps you. I've been at my computer for more than six hours today, so it's time for me to get off. perhaps we'll meet again online and continue. I hope so.

Before I "hang up,"

Rewrite the second function as a power function (drop the radical symbol). Remember to use the chain rule when finding the derivative.

The third function is a product. Use the product and chain rules, also the power rule, to obtain the derivative.

Answer 13

for \[\sqrt{2x}\] i got
f'(x)=(1/2)( 2x ^{1/2-1}
f'(x)=2x ^{-1/2}
f'(x)=1/2x\[f'(x)=1/2x^{1/2}\]
is it right?

Answer 14

for the 3rd one I got \[f'(x)=10x ^{4}-20^{3}-20x+8\]

Answer 15

\[f'(x)=40x ^{3}-60x ^{2}-20\]

Answer 16

Thanks for your efforts. This problem is a bit sticky because we have to use parentheses and use them exactly right to obtain the correct derivative.

If you know how to use the Equation Editor, I'd suggest using that.

If not, or if you'd prefer not to, then here goes:

f(x)= Sqrt(2x) = (2x)^(1/2) (Note how I've used parentheses around both 2x and 1/2).

Then the derivative f'(x) will be (1/2) * (2x)^[(1/2)-1] * derivative of (2x).

The same thing, in the equation editor, is
\[f'(x)=\frac{ 1 }{ 2 }(2x)^{(\frac{ 1 }{ 2 }-1)}*2,\]
where that final 2 is the derivative of (2x) (chain rule).

Simplifying, \[f'(x)=(2x)^{(-1/2)},\]

Answer 17

or \[f'(x)=\frac{ 1 }{ \sqrt{2x}}\]

Answer 18

oh so i missed out on the derivative of 2x at the end

Answer 19

Yes. There are other significant differences as well. I really need to get off my computer now, but encourage you to follow my explanation as best you can and to message me if you need further explanations. All the best to you. Good night.

Answer 20

thx for your help. hope to see u again

Answer 21

is the answer for the 3rd function right?

Answer 22

Hi! I calculated the derivative for the third problem and got the same result as you, EXCEPT that your result doesn't have an x^2 term. I got 18x^2.

But you're so close, I'd suggest you just move on to other problems. You're doing fine!

Answer 23

18x^2?? how?