Question

Find the derivatives of the following using implicit differentiation.
x=siny+cosx

Answer 1

Hello and Welcome to OpenStudy! :)

what's the derivative of sinx = ?
cosx = ?
and x = ?

Answer 2

hi, actually i alrady did it just wanna make sure is this dy/dx=1+sinx/cosy what u get?

Answer 3

yeah :)
just remember the parenthesis
dy/dx = (1+sinx)/cosy

Answer 4

can i ask u some more questions ?

Answer 5

sure :)

Answer 6

Find the derivatives of each function, and simplify as much as possible

y=x/2-sin2x/4

Answer 7

could u just help me with first few steps cuz when they get sepratd like that i get confused

Answer 8

when separated by addition or subtraction you can just take the derivative separately
so like

x/2 -> ?
and
(sin(2x))/x -> ?

Answer 9

1 and 2cos2x??

Answer 10

btw its nots sin2x/x its sin2x/4

Answer 11

yeah typo sorry :P

mmm try the second one again?
then chain rule should bring out a 2
which can multiply with the /4 part

Answer 12

ohh do we have to use the quotint rule on each on the right?

Answer 13

you can if you want 0.o

but (sin(2x)) /4 = (1/4) (sin(2x))
so you just find the derivative of sin(2x)
then times (1/4)

Answer 14

is it easier to do it ur way right

Answer 15

yep :) because as long as you don't have a variable in the denominator
you don't need quotient rule :P

Answer 16

soo its cos2x/2?

Answer 17

so does the x/2 become 1/2/?

Answer 18

yep for that part :)

then you the two parts back together :)

Answer 19

so does the x/2 become 1/2/?

Answer 20

yep :)

Answer 21

so the final answer is 1-cos2x/2?

Answer 22

yeah (1-cos2x)/2

Answer 23

wow thanks man

Answer 24

can i ask more ?

Answer 25

of course :)

Answer 26

Find the derivatives of each function, and simplify as much as possible

f(x)=sin(5x)/cos(x^2) , ok so for this i got as my final answer 10cos(x^2)cos(x)+sin(5x)sin(x^2)/cosx^4, is it correct??

Answer 27

mmm that's not what I got?

Answer 28

yea i had a feeling i was wrong

Answer 29

quotient rule...
http://www.math.hmc.edu/calculus/tutorials/quotient_rule/

and (cos(x^2))^2 does not equal cos(x^4)
it equals
\[\cos ^{2}(x ^{2})\]

Answer 30

its not cos(x^2)^2 its just cos(x^2)

Answer 31

yeah but in the denominator, after the quotient rule you'll get a cos(x^2)^2

Answer 32

ohhh yeaa

Answer 33

is that the only thing i did wrong ??

Answer 34

some of the chain rule wasn't right on top.
and the two parts are subtracted not added

f(x) (g(x))(f'(x)) - (f(x))(g'(x))
--- = --------------------
g(x) (g(x))^2

Answer 35

ok so i redid it and his is my final soultion , 5cos(x^2)cos(5x)+2sin(5x)(sinx^2)/cos^2x^2 , am i right now ??

Answer 36

chain rule from cos(x^2) should give you a 2x right?

Answer 37

oohh yeaa ur right

Answer 38

so it become -2sin2x??

Answer 39

more like 2x(sin(x^2)) somewhere

Answer 40

but the derivative of cos is -sinx

Answer 41

yeah sorry, my brain's fried XD

Answer 42

so its -2xsinx^2 right

Answer 43

yep :)
so
5cos(x^2)cos(5x)+2xsin(5x)(sinx^2)/cos^2x^2

Answer 44

but woulnt sin(5x) be 5sin(5x)??

Answer 45

i mean 5cos5x

Answer 46

5cos(x^2)cos(5x)+2xsin(5x)(sinx^2)/cos^2x^2
^ ^ like there?

Answer 47

yea

Answer 48

ohh yea nvm

Answer 49

can i ask more ?

Answer 50

sure :)

Answer 51

same topic , y=3sin^4(2-x)^(-1) , ok so on this one i was completly stuck no idea what to do next

Answer 52

like
\[3~\sin ^{2}(2-x)^{-1}~~ or ~~ (3 \sin ^{2}(2-x))^{1}\]

Answer 53

gah typo on the second ^(-1)

Answer 54

its sin^4 not ^2

Answer 55

yeah okay, but which one?

Answer 56

in my question

Answer 57

no, I mean how does the ^(-1) carry?

Answer 58

like u mean what is the next step?

Answer 59

like is it (3sin^4(2-x))^(-1) or 3sin^4(2-x)^(-1) ?

Answer 60

2nd one

Answer 61

well if it's 3sin^4(2-x)^(-1)
then it can be looked at like 3 (sin(2-x))^(-4)

Answer 62

oh ok

Answer 63

then chain rule and chain rule and simplify...

Answer 64

i dont know what to do with sin^4(2-x)

Answer 65

chain rule...
3 (-4) (sin(2-x))^(-5) (chain rule)

Answer 66

wouldnt sin^(4) become sin^(3)

Answer 67

there was a ^(-1) somewhere
so (sin^4 (2-x))^(-1) becomes (sin(2-x))^(-4)

Answer 68

after that i get -12sin(2-x)

Answer 69

? derivative of sin is cos?

Answer 70

so -12cos(2-x)??

Answer 71

I got that 3 (sin(2-x))^(-4)
-> (12 (cos(2-x)) (sin(2-x))^-5 )

Answer 72

what happens with 2-x?

Answer 73

huh?

Answer 74

dont we do somthing with the 2-x that is inside the sin

Answer 75

chain rule
so derivative of (2-x) is -1
which canceled out a negative somewhere else and made 12 positive

Answer 76

ohh

Answer 77

so as my final solution i get 12cos(2-x)sin(2-x)^(-5) , am i correct

Answer 78

yep :)
and if you don't like the negative exponent you can make it

(12(cos(2-x)) / (sin^5(2-x))

Answer 79

thanks

Answer 80

you're welcome :)

Answer 81

can we do some more plz?

Answer 82

Use a calculator to approximate the slope of the tangent line drawn to the graph of
y = 2sin x + cos x at x = 4.2 . Round your answer to two decimal places, i dont understand what the question really means could u explian to me

Answer 83

find the derivative
then plug in x=4.2 into the derivative and solve

Answer 84

ok i get 1.92 right?

Answer 85

make sure you're in radians mode :)

Answer 86

so -0.11?

Answer 87

and how did u know we have to be in radians mode ?

Answer 88

yep :) and
calculus = radians mode (because of the pi stuff)
(if mentions degrees) = degrees mode

Answer 89

so everything i do in calculus it has to be in radains ? or does that only apply to trigomentric dervativcs?

Answer 90

calculus.

Answer 91

what do they mean here Write the equation of the tangent line of 2cosx at x=-pi/2

Answer 92

find the derivative
plug in the x value into the derivative and solve

plug in the x value into the original equation and solve

use the point and the slope to make the equation into point slope form

Answer 93

ok for the first 2 steps i get 2 and -0.832293673 , i dony understad what do to in thr thisrd step

Answer 94

I got 0 for the second
2cos(-pi/2) = 0

then point slope form is
y - y1 = m(x - x1)

so plug in m = 2
x1 = - pi/2
and y1 = 0

Answer 95

in the second one isnt it -2sin(-pi/2)? like when u sub it in after u take the derviative

Answer 96

mmm
my point was to find the y coordinate part
from the original equal to get a (-pi/2, something)

then use the derivative -2sinx
plug in for x to get slope = m

Answer 97

y = 2x + pi is what I got

I've gtg now,
If you still need more help,
you can tag people by putting a @
in front of their username like @hughfes
some helpers that can help you are
jdoe0001
whpalmer4
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good luck! :)