Question

Compute (d/dx) [sin(cos(x))]

Answer 1

please post all calc involved

Answer 2

use chain rule
u = cos x
du = -sin x

d/du sin(u) = cos(u)

--> d/dx sin(cos) = -sin(x)*cos(cos(x))

Answer 3

Yep,
(sinu)' = u' cos u = -sinx * cos (cosx)

Answer 4

does the same apply for

(d/dx)[tan(tan(x))]

Answer 5

(tan u)' = u' sec^2 u

Answer 6

u = tanx --> u' = sec^2x

Answer 7

ok i got. one more thing.

when it says compute y' for x^2+sqrt(xy)=xy^5

am i calculating derivative with respect to y

Answer 8

Yes

Answer 9

2x + ( y + xy')/ 2 sqrt (xy) = y^5 + 4xy^4 y'

Answer 10

is that the final?

Answer 11

can you help me with another one thats quite similar.

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

Answer 12

( y + xy') cos(x+y) = 2x + 2yy'

Answer 13

xy' cos(x+y) - 2yy' = 2x - ycos(x+y)
y' (xcos(x+y) - 2y) = 2x - ycos(x+y)
=> y' = [2x - ycos(x+y)] / [xcos(x+y) - 2y]

Answer 14

ok were you going to add anything about the previous question. i seem to have posted this new question right when you were about to type something

Answer 15

I have the hard time to look at it on screen, I think you should post it to get second opinion!

Answer 16

\[x^2 +\sqrt{xy}=xy^5\] compute y'

i will post it to get a second opinion though. thanks for the help

Answer 17

y' = 2√(xy) ( y^5 - 2x) / [ x - 8xy^4 √(xy)

Answer 18

Good luck, Jinnie :)

Answer 19

thank you

Answer 20

I'll check back on you frequently :)

Answer 21

alright. thanks once again