Question

Use implicit differentiation to find an equation of the tangent line to the curve.
sin(x+y)=6x−6y at the point (π,π).

Answer 1

i would like it if you told me the steps and then check my steps after i do them. thanks

Answer 2

Use chain rule to find the derivative of the first term.

Answer 3

so the inner function would be x+y and the outside would be sinx?

Answer 4

Derivative of sin( something ) = cos( something) times derivative of (something)

Answer 5

would the derivative of x+y=0

Answer 6

derivative of x with respect to x is 1
derivative of y with respect to x is dy/dx. But it is easier in implicit differentiations to write dy/dx as y'

Answer 7

ok

Answer 8

So differentiate each term in the original equation first.

Answer 9

cos(x+y)(1+y')=5-6y'?

Answer 10

Yes but the 5 on the right should be 6.

Answer 11

oh yeah sorry

Answer 12

now i have cos(x+y)+y'cos(x+y)=6(1-y')

Answer 13

{cos(x+y)} (1+y') = 6 - 6y'

There are two terms with y'. Group them together and express it as y' = something

Answer 14

ok so cos(x+y)-6=6y'-y'cos(x+y)

Answer 15

sorry -6y

Answer 16

-6y'

Answer 17

yeah sorry i cant type right now

Answer 18

factor y' out on the right and express it as y' = something

Answer 19

ok so it would be y'=\[\cos (x+y)-6/6+\cos(x+y)\]

Answer 20

would that be -1

Answer 21

yes there should be a -outside or you can switch the numerator to: 6 - cos(x + y)

Answer 22

ok so would that equal -1

Answer 23

would what equal -1? Answer?

Answer 24

umm yes?

Answer 25

You get y' = { 6 - cos(x + y) } / { 6 + cos(x + y) }

y' is the slope of the tangent line. Evaluate the slope at x = pi and y = pi

Once you know the slope and a point on the line (pi, pi) you can find the equation of the tangent line at (pi, pi).

Answer 26

is the slope 5/7

Answer 27

Yes.

Answer 28

so i do the point slope form with (pi,pi) as the point?

Answer 29

Yes.

Answer 30

ok i got y=(5/7)x+(2/7)pi

Answer 31

Yes. Does it agree with the answer. You may have to put in the value for pi.

Answer 32

yup its right. thanks

Answer 33

You are welcome.

Answer 34

@ranga another problem i had was 2x^2+2xy+y^2=5 at the point (1,1). i got y'=-2x/(y+1) is that right?

Answer 35

I am getting y' = -(2x + y) / (x + y)

Answer 36

oh ok. how

Answer 37

2x^2+2xy+y^2=5
Differentiate each term:
4x + (2xy' + 2y) + 2yy' = 0
4x + 2y + 2xy' + 2yy' = 0
2(2x + y) + 2y'(x + y) = 0
(2x + y) + y'(x + y) = 0
y' = -(2x + y) / (x + y)

Answer 38

ok i did the 2xy derivative wrong

Answer 39

You got to apply product rule.

Answer 40

ok i got it right now thanks

Answer 41

Glad to help.

Answer 42

what would the derivative of y^2-4 be? 2yy'?

Answer 43

yes.

Answer 44

and x^2-5=2x?

Answer 45

yes.

Answer 46

ok thanks

Answer 47

sure.

Answer 48

would the derivative of 5+x^4y=4x^3y+x^4y'?

Answer 49

and what about arctan(xy)=1/(1+(xy)^2)

Answer 50

@ranga

Answer 51

The principle is the same as before. There are just a few rules for finding the derivative: power rule, chain rule, product rule, quotient rule, etc.

I will take just one term from your equation and differentiate: x^4y
x^4y' + 4x^3y'

For the last term: if arctan(A) = B, then it means A = tan(B)

Answer 52

i dont get it

Answer 53

i am asking if u if the derivative of 5+x^4y=4x^3y+x^4y'? also if the derivative of arctan(xy)=1/(1+(xy)^2)

Answer 54

Oh, I thought that was the original equation they gave.

Yes the derivative of 5+x^4y is 4x^3y+x^4y'

Give me a minute for the other one.

Answer 55

derivative of arctan( something ) = 1 / { 1 + (something)^2 } times derivative of ( something )

Answer 56

ok

Answer 57

so derivative of arctan(xy)=(1/(1+(xy)^2))(y')

Answer 58

The 1 / { 1 + (xy)^2 } part is correct. But the derivative of xy: xy' + y

Answer 59

ok

Answer 60

alright.

Answer 61

i tried to figure it out myself but i cant arctan(xy)=5+x^4y

Answer 62

i just need to find the derivative

Answer 63

@ranga

Answer 64

You need derivative of arctan(xy) ?

Answer 65

i need the derivative of the whole thing

Answer 66

Is the question: find dy/dx or y' ?

Answer 67

arent they the same thing

Answer 68

well its the same as before

Answer 69

They are the same thing. But I am asking what is the question? You can differentiate the right hand side and you can differentiate the right hand side. That will just leave some expression on the right = some expression on the right. But what are they asking you to find?

Answer 70

y' of the whole thing

Answer 71

Okay that makes sense.

Differentiate the left side. We just did the arctan thing earlier.
Differentiate the right side.

Group all y' terms and then express it as y' = some expression

Answer 72

it doesnt specify whether we have to use implicit differentiation or not. but i got 1/(1+(xy)^2)(xy+y')=4x^3y+x^4y

Answer 73

\[1/(1+(xy)^2)(xy'+y)=4x^3y+x^4y\]

Answer 74

The last term should be x^4y'

Answer 75

ok

Answer 76

and is it (xy+y') or (xy'+y)

Answer 77

at the beginning

Answer 78

derivative of xy is: xy' + y (just product rule: first term times derivative of second term + second term times derivative of the first term)

Answer 79

ok

Answer 80

so how would i do it

Answer 81

(xy' + y) / (1+ (xy)^2) = 4x^3y + x^4y'

Multiply throughout by (1+ (xy)^2)
Group y' terms
Express it as y' = some expression

Answer 82

it would be (1+x^2y^2)

Answer 83

how would u group the y'

Answer 84

Instead of typing the whole thing which is time consuming I am going to call
(1+ (xy)^2) as t
(xy' + y) / t = 4x^3y + x^4y'
Multiply throughout by t
(xy' + y) = 4tx^3y + tx^4y'
xy' - tx^4y' = 4t^3y - y = y(4t^3 - 1)
y'(x - tx^4) = y(4t^3 - 1)
y' = y(4t^3 - 1) / (x - tx^4)
put t back as (1+ (xy)^2)

Answer 85

y(4(1+(xy)2)3−1)x−(1+(xy)2)x4 this is what i plugged in and i got it wrong

Answer 86

If I expand all the terms and then factor it out I get:\[\Large y' = \frac{ y(-4x ^{5}y^{2} - 4x^{3} + 1) }{ x(x^{5}y^{2} + x^{3} - 1) }\]

The method is correct. But if the answer doesn't agree it must be some trivial error and rechecking will catch the error.