# Derivative of square root

1. Feb 4, 2013

### Torshi

1. The problem statement, all variables and given/known data

Finding derivative of (sin(sqt3x+5))

2. Relevant equations

None given. Chain Rule

3. The attempt at a solution

The answer is: (cos(sqrt3x+5)) * 1/2(sqrt3x+5) * 3

but I don't know how to get to the 3.

I turned sin into cos and multiplied by the inside derivative giving the first two parts of the answer. How do I get 3? - is that from the inside derivative of 3x+5? How do you know which is outside or inside of a derivative

2. Feb 4, 2013

### Torshi

Another problem:

5√(x^2+1)^4 <-- That's all under the square root.

i know the answer. But how do I figure out which is the outside then multiply that by the inside derivative?

I understand how this becomes [(x^2+1)^4]^1/5 but then it turns into (x^2+1)^4/5 were the exponents multiplied 4/1 * 1/5 = 4/5?

Then from there it becomes 4/5(x^2+1)^-1/5 * 2x

Last edited: Feb 4, 2013
3. Feb 4, 2013

### Dick

sin() is outside, sqrt() is inside of that. And 3x+5 is inside of sqrt(). You need two stages of chain rule.

4. Feb 4, 2013

### Dick

5*sqrt() is outside. ()^4 is inside of that. And inside of that is x^2+1

5. Feb 4, 2013

### Torshi

√x^3tan(x) <-- all under sqrt.

so I did...

1/2*(x^3tan(x))^-1/2 * 3x^2sec(x)^2

but.. the answer says 1/2√x^3tan(x) * (3x^2tan(x) + x^3(sec(x))^2)

I do know that √u = 1/2√u, but I wrote 1/2*(x^3tan(x))^-1/2 due to exponent rule?

6. Feb 4, 2013

### Dick

That's just fine until you tried to find the derivative of x^3*tan(x). Now you need to use the product rule.

7. Feb 4, 2013

### Torshi

Haha thank you! I stumbled for sure.

8. Feb 4, 2013

### Torshi

Quick question:

there is a problem i did which was (sin(x))^2

= 2cos(x)) * 1 = 2cos(x) using chain rule, but for the problem I had to use the product rule in which I got...

(sin(x))^2
=(sin(x)(sin(x))
=(cos(x))(sin(x))+(sin(x))(cos(x))
= 2(sin(x))(cos(x))

do both answers equal each other? Trig identity unless I messed up somehow?

Last edited: Feb 4, 2013
9. Feb 4, 2013

### Dick

No. They aren't the same. The derivative of u^2 is 2*u*u'. The chain rule gives you 2*sin(x)*(sin(x))'=2*sin(x)*cos(x). NOT 2*cos(x). Same as the product rule.

10. Feb 4, 2013

### Torshi

So the exponent 2 comes down multiplying with sin(x) then sin(x) is considered the inside derivative then you multiply that by cos(x)?

I don't understand why it just doesn't become 2cos(x)

11. Feb 5, 2013

### Dick

Yes! Yes! "the exponent 2 comes down multiplying with sin(x)" then you multiply by the derivative of sin(x). What happen to the sin(x) you multiplied by 2? Did you just drop it? This is just being sloppy.

12. Feb 5, 2013

### Torshi

Haha alright it's just I have a hard time differentiating between inside and outside so 2*sin(x) with sin(x) turning into 2*cos(x) looks done, but I guess not. Thank you though!

13. Feb 5, 2013

### Staff: Mentor

Someone reading your work would have no idea what you are doing, since you give no indication that you are taking a derivative.

(sin(x))2 = sin(x) * sin(x), but
sin(x) * sin(x) cos(x) sin(x)+ sin(x)cos(x)

What you should be saying is that
d/dx[sin(x) * sin(x)] = cos(x) * sin(x) + sin(x) * cos(x) = 2 * sin(x) * cos(x)

You could have gotten this same result by using the chain rule when you differentiate sin2(x).