Question

Find the equation of the tangent line to the curve
y=sin(sin(5x))+e^x^2

at the point where x=0. Your answer must be in the form of an equation for y in terms of x, that is, of the form y=L(x) for some linear function L(x).

Answer 1

derivative is (cos(cos(5x) 5) +e^x^2 (2x)

Answer 2

e^(x^2)
or
(e^x)^2
??
these 2 are different

Answer 3

the first one

Answer 4

your derivative of sin sin 5x is incorrect

Answer 5

heard about chain rule ?

Answer 6

yes

Answer 7

i got the derivative at the top

Answer 8

i think it's right

Answer 9

\((sin (sin 5x))' = cos (sin 5x) \times (sin 5x)' \)

Answer 10

oh yeah

Answer 11

my mistake

Answer 12

whats your new derivative ?

Answer 13

cos(sin5x)×(sin5x)′ + e^x^2 (2x)

Answer 14

cos(sinx5x) * (sin5x)' +e^x^2 2x)

Answer 15

(sin 5x)' means derivative of sin 5x

which is \( \cos 5x \times 5 \)

Answer 16

5cos(sinx5x) * (cos 5x) +2x e^x^2

Answer 17

okay then now i just have to make it equal to 0?

Answer 18

thats y'
plug in x=0 to get the slope at that point

Answer 19

okay?

Answer 20

then i get the slope?

Answer 21

yup

Answer 22

wouldn't i get 0 for the slope?

Answer 23

nope

Answer 24

cos 0 =1

Answer 25

so the slope would be 1?

Answer 26

no 5

Answer 27

bingo!
5 is correct :)

Answer 28

now what do i do with the 5?

Answer 29

you have slope
you have x co-ordinate = 0
what else do u need to get the equation of the line ?

Answer 30

i mean x co-ordinate of a point

Answer 31

b?

Answer 32

you know you get get the equation of line if you have slope and a point

Answer 33

not really

Answer 34

yoriginal - y = m(xoriginal - x) ?

Answer 35

then solve for y?

Answer 36

you need y co-ordinate
find y when x=0

plug in x=0 in the original equation

Answer 37

the answer would be 1 then right?

Answer 38

yes

so y co-ordinate is 1
point is (0,1)
slope is 5
find the equation

Answer 39

the answer would be y=5x+1

Answer 40

thanks, again!

Answer 41

welcome ^_^