Question

Find Devivaties with respect to x:

1). y=sin(cos(x+3));

2). tan(x+y)=ln x+ 5y.

Answer 1

1. - use the chain rule

dy/dx = cos (cos(x+3) * sin (x + 3)

Answer 2

i thought so but then i was thinking product rule...this is a review question for my calc 2 class

Answer 3

chain rule: - look for a 'function within a function'

Answer 4

1). y=sin(cos(x+3));

2). tan(x+y)=ln x+ 5y.

1:y = sin(cos(x+3))
y' = cos(cos(x+3))(-sin(x+3))(1)
y' = {-sin(x+3)}{cos(cos(x+3))}

2. Tan(x+y) = lnx+5y
Sec^2(x+y)(1+y') = 1/x + 5yy'
y' = {((1/x))-(Sec^2(x+y))}/(Sec^2(x+y) - 5y)

My calculus is rough right now. 2 im not sure but 1 im sure

Answer 5

product rule is used for product of 2 or more functions

Answer 6

oh yes - i made a mistake on no. 1:- the derivative of cos is - sin
franciscanmpnk is correct

Answer 7

yeaa i caught that ...his number 2 seems correct as well ...Thanks to the both of you!

Answer 8

yw

Answer 9

for no 1:
{-sin(x+3)}*{cos(cos(x+3)}

Answer 10

yes your correct...how about number 2...need a second opinion

Answer 11

no 2 is implicit differentiation where y is taken to be a function of x
tan(x+y)=ln x+ 5y.

use chain rule for tan(x + y)
= sec^2(x + y) * (1 + y')
ln x + 5y = 1/x + 5yy'
sec^2(x + y) * (1 + y') = 1/x + 5yy'
now we need to make y' the subject

Answer 12

make sense

Answer 13

makes sense i mean

Answer 14

yes cw, that is that is just what i would do.

Answer 15

sec^2(x + y) + y' sec^2(x + y) = 1/x + 5yy'
y' sec^2(x + y) - 5yy' = 1/x - sec^2(x + y)
y' = [ 1/x - sec^2(x + y) ] / [sec^2(x + y) - 5y]

Answer 16

- same as franciscanmonk got

Answer 17

thaks cw...you explain things very clear

Answer 18

ty tvvv1012 but there is a mistake in the above
the derivative of 5y is NOT 5yy' but 5y'

Answer 19

so both francis and i are wrong

Answer 20

y' sec^2(x + y) - 5y' = 1/x - sec^2(x + y)
y' = [ 1/x - sec^2(x + y) ] / [sec^2(x + y) - 5]

Answer 21

haha sorry about no.2

Answer 22

i made the same mistake at first - hey - shows we're human!!!