Question

f(x)=5xsinxcosx,
f'(x)=
f(x)=1sinxcosx
f'(x)=

Answer 1

do you know product rule?

Answer 2

yes

Answer 3

okay let's use product rule for 3 variables.
(f(x) g(x) h(x))' = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)

Answer 4

right,

Answer 5

so just plug the numbers into the equation

Answer 6

no ... use f(x) = 5x , g(x) = cos(x) and h(x) = sin(x) on the above formula

Answer 7

ok,
(5x)'(sinx)(cosx)+(5x)(sinx)`(cosx)+(5x)(sinx)(cosx)' ?

Answer 8

I got (5)(cosx)(sinx)+(5x)(cosx)(cosx)+(5x)(sinx)(sinx)

Answer 9

this should be
+(5x)(sinx)(sinx) = -(5x)(sinx)(sinx)

Answer 10

ops thank you.

Answer 11

for the next one how do i do?

Answer 12

do same for next one ... just use f(x) = 5

Answer 13

sorry .. 1

Answer 14

ok, i will wrok on it

Answer 15

i got
(sinx)(cosx)+(cosx^2)-(sinx)^2

Answer 16

No ... it should be like this
1 sinx cosx = cos^2(x) - sin^2(x)

Answer 17

how did you get this?

Answer 18

you didn't use the formula i gave you correctly ... did you?

Answer 19

ops, you are right,
i have another questions

Answer 20

\[\lim_{x \rightarrow 0} tanx/2x\]
sin3x/5x

Answer 21

it says Evaluate the limit

Answer 22

sec^2/2x?

Answer 23

do you know that lim x-> 0 sin(x)/x = 1 ?

Answer 24

Yes,

Answer 25

write tan(x) as sin(x)/cos(x) ... and then separate it as above.

Answer 26

sinx/cosx*2x?

Answer 27

separate the limits like this

Now evaluate the limit individually and multiply them.