Question

find the equation of the tangent line to the graph of y = x + tanx at the point (pi , pi). write in slope intercept form

Answer 1

i used the first derivative test and i got 1+sec^2(x)

Answer 2

idk what to do next

Answer 3

find the slope by pluggin in pi

Answer 4

into your derivative

Answer 5

so in the calculator it would look 1+ 1/(cos(pi)^2 right? i got 2 for the slope

Answer 6

cos(pi)=-1
cos^2(pi)=1
so 1/cos^2(pi)=1/1=1
so yeah 1+1 is 2

Answer 7

the slope if 2

Answer 8

y-y1=m(x-x1)

Answer 9

you know m=2

Answer 10

and you also can find a point on the curve and that is on the tangent line

Answer 11

yeah i got y-pi= 2(x-pi)

Answer 12

oh yeah it was already given
was thinking you had to plug in pi
just to figure out it was pi as output too :)

Answer 13

:D thanks

Answer 14

now put that in whatever form you want and can
but that is the tangent line in point slope form

Answer 15

wait

Answer 16

so for example if in the quiz they ask me the same thing but for a different question like y= x+cotx

Answer 17

i just use the first derivative

Answer 18

and evaluate for pi right?

Answer 19

if (pi,pi) was given like this time

Answer 20

well if it says to find the tangent line at x=pi yeah

Answer 21

but cot(pi)=cos(pi)/sin(pi)
which doesn't exist

Answer 22

so that question actually wouldn't make sense

Answer 23

hehe :D

Answer 24

you know what i meant though. thanks alot

Answer 25

finish what you are typing i got a last question :)

Answer 26

if we are asked to find the tagent line to y=x+cot(x) at (a,b)
then first step
find slope=1-csc^2(x)
find slope (@x=a)=1-csc^2(a) <--this is slope
so we have at x=a, y=a+cot(a)=b
so the tangent line here would be
y-b=m(x-a)
y-a-cot(a)=(1-csc^2(a))(x-a)
y=(1-csc^2(a))(x-a)+a+cot(a)
y=(1-csc^2(a))x-a(1-csc^2(a)+a+cot(a)
y=(1-csc^2(a))x-a-acsc^2(a)+a+cot(a)
y=(1-csc^2(a))x-acsc^2(a)+cot(a)

Answer 27

there is a question in my practice quiz like this y=sin(sin(sinx)

Answer 28

find y'?

Answer 29

no differentiate

Answer 30

?

Answer 31

don't find y'?

Answer 32

or do?

Answer 33

it says differentiate that function

Answer 34

so that means we need to find y'

Answer 35

or to find the derivative

Answer 36

yep i wanted to know if there was a difference between that function and sinx.sinx.sinx

Answer 37

\[y=f(u(x)) \\ y'=f'(z)|_{z=u(x)} \cdot u'(x)\]
do you know the chain rule?

Answer 38

yes

Answer 39

yes sin(x)*sin(x)*sin(x) is a product

Answer 40

sin(sin(sin(x)))
is a function inside a function inside a function

Answer 41

hmm

Answer 42

one requires product rule
the other requires chain rule

Answer 43

they are not the same function at all

Answer 44

so for the product one would you get this

Answer 45

cosxsinxsinx+ sinxcosxsinx+sinxsinxcosx?

Answer 46

for example at x=pi/2
sin(pi/2)*sin(pi/2)*sin(pi/2)
=1*1*1=1

but sin(sin(sin(pi/2)))
=sin(sin(1))
approx sin(.84147)
approx .745623

but anyways...
and yes to that question about the product or you could write it as sin^3(x)
and then differentiate
either way
\[\frac{d}{dx}\sin^3(x)=3\sin^2(x)\cos(x) \]
that is exactly what you get doing it the way you chose

Answer 47

i see thanks :D

Answer 48

but the chain rule one can you do that one?

Answer 49

would that be cos(sin(sinx).cosx(sinx).cosx

Answer 50

\[y=f(u(x)) \\ y'=f'(z)|_{z=u(x)} \cdot u'(x) \\ y=\sin(\sin(\sin(x)))\]
let f(z)=sin(z) and u(x)=sin(sin(x))
so f'(z)=cos(z)
so f'(u(x))=cos(sin(sin(x))
and
we could use the chain rule again for the sin(sin(x)) part
so u(x)=g(h(x))
u'(x)=g'(h(x))*h'(x)
u'(x)=cos(sin(x))*cos(x)
so yeah
you get
y'=cos(sin(sin(x))*cos(sin(x))*cos(x)

Answer 51

good job

Answer 52

i know that is probably longer of an explanation then you wanted
just trying to be thorough

Answer 53

that's perfect thanks