Can You Solve These Calculus Derivative Problems Correctly?

[Alethia]

Well I was doing my calculus homework today and I got stuck one some problems. Well actually I just wanted to check if I did them right because I'm not entirely sure. Anybody care to check? :D

1) y = sin (9[x^(1/2)] + sin [x^(1/2)]
--For this problem, I just used the chain rule for each part. For my answer I got [cos(x^91/2)]/[x^(1/2)]. I wasn't sure if I could do the chain rule for both of them and just add them. Was that okay?

2) y = 4sec^2 x(tan x)
--In this problem, I used the product rule then the chain rule inside. I think I did it right but the 4 kind of threw me off. My final answer was: 4sec^2x (sec^2x + 2tan^2x).

Thanks! Anythign helps really... :D

 

[NateTG]

It's much easier to help if you show work. That said:

1.The way you've written it, the parens don't balance, but in general, the derivative of a (finite) sum is the sum of the derivatives. Also your answer doesn't look right to me I think you did some bad algebra pulling things out of the trig functions.

2. It seems like there should be an extra 2 from the $$\sec^{x}$$ in the derivative that you're missing.

 

[Alethia]

HAha yeah it would be easier that way wouldn't it? Okay here's my work, that way I can pin point wher eI made a mistake, apparently. :D

1) y = sin x^1/2 + sin x^1/2
y' = (cos x^1/2)(1/2x^-1/2) + (cos x^1/2)(1/2x^-1/2)
y' = (2cosx^1/2)/(2x^1/2 )
y' = cos x^1/2 /x^1/2

For the second problem...
y = 4sec²x(tan x)
y' = tan x(8 sec x(sec x * tan x)) + 4(sec x)²(sec²x
y' = tan² x(2)) + 4(sec x)²(sec²x
y' = 4 sec²x[sec² x +2 tan² x]

I hope that helps as much as it may help me. :D

 

[NateTG]

1) y = sin x^1/2 + sin x^1/2
y' = (cos x^1/2)(1/2x^-1/2) + (cos x^1/2)(1/2x^-1/2)
y' = (2cosx^1/2)/(2x^1/2 )
y' = cos x^1/2 /x^1/2

The way you originally posted it, it looked more like:
$$y=\sin(9x^{\frac{1}{2}}) + \sin(x^{\frac{1}{2}})$$
which would lead to a different derivative. The stuff above all looks good.

For the second problem...
y = 4sec²x(tan x)
y' = tan x(8 sec x(sec x * tan x)) + 4(sec x)²(sec²x
y' = tan² x(2)) + 4(sec x)²(sec²x
y' = 4 sec²x[sec² x +2 tan² x]

Also looks good. Here's a quick check:
$$y=4 \sec^2x\tanx=4\frac{\sin x}{\cos^3 x}=4 \sin x \cos^{-3} x$$
$$y'=4(\cos x \cos^{-3} x + \sin x (-3) \cos^{-4} x(-\sin x))$$
$$y'=4(\cos^{-2} x + 3 \sin^2 \cos^{-4})$$
$$y'=4 \sec^2x(1+3 \tan^2 x)$$
and
$$1+\tan^2=\sec^2$$
so
$$y'=4 \sec^2x(\sec^2 x + 2 \tan^2 x)$$