# Thread: Check my Derivatives =)

1. ## Check my Derivatives =)

I hope I did these right... thanks in advance!

g(x) = (3x-2)/(2x+1)^(1/2)

g'(x) = (9x-1)/(2x+1)^(5/2)

y = sin(cos(tan(x))^(1/2))

y' = [-cos^2(tan(x))^(1/2)sin(tanx)^(1/2)sec^2(x)]/2(tan(x))^(1/2)

these next ones dont need to be simplified

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

f'(x) = cos(X^(2) + x) * (2x+1)

f(x) = sin((cosx)/x)

f'(x) = cos((cosx)/x) * [ (x(sinx) - cosx)/x^2]

f(x) = sin^3(x^(2)+sinx)

f'(x) = 3[sin(x^(2) + sinx)]^2 * cos(x^(2) + sinx) * (2x + cosx)

f(x) = sin(xsinx) + sin(sin(x^2))

f'(x) = cos(xsinx) * [x(cosx) + (sinx)] + cos(sin(x^(2)) * cos(x^(2)) * 2x

heres the last one...

Find an equation of the tangent line to the curve f(x) = sin(sinx) at the point (pi, 0)

so i found:

f'(x) = cos(sinx)cosx

i plugged that into the y=mx+b as the slope of the tangent line and I got

cos(sin(pi)) * cos(pi) =-1

0=(-1)(pi) + b

b= pi

so, y= -x + pi

Thank you so much for checking this out for me =) I will definitely try to contribute myself around here.

2. Originally Posted by drain
I hope I did these right... thanks in advance!

g(x) = (3x-2)/(2x+1)^(1/2)

g'(x) = (9x-1)/(2x+1)^(5/2)
You can integrate it to determine if your derivative is correct.

For the first one, I got the following:

g(x) = (3x - 2)/(2x + 1)^(1/2)
g'(x) = [3*(2x+1)^(1/2) - [(1/2)*(2x + 1)*2*(3x - 2)]]/2x + 1

3. Originally Posted by drain

y = sin(cos(tan(x))^(1/2))

y' = [-cos^2(tan(x))^(1/2)sin(tanx)^(1/2)sec^2(x)]/2(tan(x))^(1/2)
This one gets quite complicated, and I'll leave you to simplify it.

y = sin(cos(tan(x))^(1/2))
y' = cos(cos(tan(x))^(1/2))*(1/2)*[cos(tan(x))]^(-1/2)*-sin(tan(x))*sec^(x)^2

4. Originally Posted by AfterShock
This one gets quite complicated, and I'll leave you to simplify it.

y = sin(cos(tan(x))^(1/2))
y' = cos(cos(tan(x))^(1/2))*(1/2)*[cos(tan(x))]^(-1/2)*-sin(tan(x))*sec^(x)^2
One other thing you have to realize in this prob; cos(tan(x)) is the term with the square root over it, not just tan(x).

5. Originally Posted by drain

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

f'(x) = cos(X^(2) + x) * (2x+1)
f(x) = sin(x + x^2)
f'(x) = cos(x + x^2)*(1 + 2x)

6. Originally Posted by drain

f(x) = sin((cosx)/x)

f'(x) = cos((cosx)/x) * [ (x(sinx) - cosx)/x^2]
f(x) = sin((cos(x))/x)

Your derivative is almost right; you have a sign error. It should be:

f'(x) = cos(cos(x)/x)*[(-sin(x)/x) - cos(x)/x^2]

7. Originally Posted by drain
f(x) = sin^3(x^(2)+sinx)

f'(x) = 3[sin(x^(2) + sinx)]^2 * cos(x^(2) + sinx) * (2x + cosx)
Wow, how many of these do you have. I'll do this last one for now.

f(x) = (sin(x^(2) + sin(x)))^3

f'(x) = 3*sin(x^2 + sin(x))^2*cos(x^2 + sin(x))*(2x + cos(x))

Yup, it looks like your derivative is what I have.

8. Originally Posted by AfterShock
One other thing you have to realize in this prob; cos(tan(x)) is the term with the square root over it, not just tan(x).
It was just the tan(x) I think I typed it it wrong =( sorry

Edit: I think I typed it right? It was supposed to be just the tan(x) in the square root.

Originally Posted by AfterShock
f(x) = sin((cos(x))/x)

Your derivative is almost right; you have a sign error. It should be:

f'(x) = cos(cos(x)/x)*[(-sin(x)/x) - cos(x)/x^2]
I had the minus sign, I mistyped again =(

Originally Posted by AfterShock
You can integrate it to determine if your derivative is correct.
I don't know what that means or how to do that.

THANK YOU VERY MUCH FOR YOUR TIME AND EFFORT. I really appreciate it, man. Thanks again.