Question

need help with attachment

Answer 1

so here is the derivative step by step but you have to apply product rule and chain rule here:
f'=3(x^1/2)*(-sin(x^1/2))(1/(2*x^1/2))+cos(x^1/2)(3/(2*x^1/2))-(1/((2*x^1/2)cos(x^1/2)
f'=(-3(x^1/2)/(2*x^1/2))sin(x^1/2)+(3/(2*x^1/2))cos(x^1/2)-(1/(2*x^1/2))cos(x^1/2)
f'=(1/x^1/2)cos(x^1/2)-(3/2)sin(x^1/2)
so the answer is number 2 on your list in word attachment let me know if you need any understanding...

Answer 2

is there anyway you could draw on the step out or do something else?

Answer 3

oops! meant draw all the steps out

Answer 4

there is no way to draw this it uses the following rules
uv=uv'+vu' where the v' and u' use chain rule here because the (X^1/2) term in those sines and cosines becomes 1/2*(x^-1/2) which is 1/(2*x^(1/2)) it just a matter of keep track of things not very hard if you look through what i wrote you should be able to pick up really fast

Answer 5

It is number 2

Answer 6

\[3x^{1/2}(-\sin(x^{1/2})(\frac{1}{2x^{1/2}})+\cos(x^{1/2})(\frac{3}{2x^{1/2}})\]

this is for your product rule before the minus sign

Answer 7

so what is your f and g in this problem?

Answer 8

cleaning it up by multiplying stuff out you'll get

\[\frac{3x^{1/2}}{2x^{1/2}}(-\sin(x^{1/2})+\frac{3}{2x^{1/2}}(\cos(x^{1/2}))\]

Answer 9

oh I see f=3x^(1/2) and g=cos^(1/2)

Answer 10

oops g=cos(x)^(1/2)

Answer 11

then doing the last part of the derivative which would be

\[\frac{d}{dx}[\sin(x^{1/2})=\cos(x^{1/2})(\frac{1}{2x^{1/2}})=\frac{1}{2x^{1/2}}(\cos(x^{1/2}))\]

Answer 12

so when you put it together you get

\[\frac{-3\sqrt{x}}{2\sqrt{x}}(\sin(\sqrt{x})+\frac{3}{2\sqrt{x}}(\cos(\sqrt{x}))-\frac{1}{2\sqrt{x}}(\cos(\sqrt{x}))\]

the square roots in the sin part will cancel and when you can minus the cos due to same denominators and youll get 2 on the top which will reduce to 1

Answer 13

LOL, my tutor would be evil next year when i learn that xD
Gawd , I hate my tutor he gives me extremely HARD h.w

Answer 14

soo your anwer after doing all in the above paragraph will be
\[\frac{-3}{2}\sin \sqrt{x}+\frac{1}{2\sqrt{x}}\cos \sqrt{x}=\frac{1}{2{\sqrt{x}}}\cos \sqrt{x}-\frac{-3}{2}\sin \sqrt{x}\]

Answer 15

which is answer 2!

Answer 16

where -3/2 sin(X)^(1/2) at the end of the function come from

Answer 17

oh, my bad its equal sign between the two equation

Answer 18

how come when you reduce the cosine function because they had common base when you did 3-1 and got 2 shouldn't that cancel out the 2 on the bottom leaving 1/x^(1/2)