Question

The twice–differentiable function f is defined for all real numbers and satisfies the following conditions:

f(0)=3

f′(0)=5

f″(0)=7

The function h is given by h(x)=cos(kx)[f(x)]+sin(x) for all real numbers, where k is a constant. Find h ′(x) and write an equation for the line tangent to the graph of h at x=0.

Answer 1

@ganeshie8 please help!!

Answer 2

will give medal :) @Loser66 @ganeshie8

Answer 3

for h'(x) , just take derivative h(x)
for tangent, replace x =0 in h'(x), what do you have?

Answer 4

How do i take the derivative of h(x)

Answer 5

what??? really?? you don't know how to take derivative of a function?

Answer 6

no :( not this function this ones complicated

Answer 7

cos(kx)[f(x)]+sin(x)

Answer 8

ok, (sin (x))' = ?

Answer 9

idk :(

Answer 10

@jim_thompson5910 what should I do?

Answer 11

what is the derivative of cosine?

Answer 12

im not sure ive been slacking so badly

Answer 13

you should have something like this in your notes

http://sub.allaboutcircuits.com/images/11045.png

Answer 14

@1019.jams We can help you cook only when you have something to cook. If you have nothing, how can we help?

Answer 15

can you please show all the steps then i'll understand from there :) that's how i learn math i always follow the steps that my teachers show me

Answer 16

@Loser66

Answer 17

@satellite73

Answer 18

1019.jams, did you look at the link I posted

Answer 19

yes; now i need to write an equation for the line tangent to the graph of h at x=0 @Loser66

Answer 20

@ganeshie8

Answer 21

so if y = sin(x), then what is dy/dx equal to?

Answer 22

cos(x) :)

Answer 23

if y = cos(x), then dy/dx = ???

Answer 24

this is what i have so far
h'(x) = [cos (x)(f(x)]' +(sin (x))'= (cos(x))' f(x) + cos(x) f'(x) + cos (x) = sin x f(x) +cos x f'(x) +cos x

Answer 25

if you are stuck at taking derivatives, this problem is not for you

Answer 26

don't forget that you had cos(kx)*f(x) and not just cos(x)*f(x)

Answer 27

so you're missing that k

Answer 28

got it :) what do i do next :)

Answer 29

how do i find an equation for the line tangent to the graph of h at x=0

Answer 30

so what is h'(x) equal to

Answer 31

h'(x) = [cos (kx)(f(x)]' +(sin (x))'= (cos(kx))' f(x) + cos(kx) f'(x) + cos (kx) = sin x f(x) +cos kx f'(x) +cos kx

Answer 32

incorrect

Answer 33

look back at the original problem, and be careful about the chain rule