Question

derivative of x^-2sin^2(x^3). how can i solve this easily?

Answer 1

use the product rule

Answer 2

so 2 at a time and then the other two?

Answer 3

as this is the product of x^-2 and sin^2(x^3)
you have to use product rule for it..

d/dx (u.v) = u . dv/dx + v . du/dx

Answer 4

so like derivative x^-2 times sin^2(x^3) + x^-2 times derivative of sin^2(x^3)?

Answer 5

yes..

Answer 6

so is it -2x^-3 times sin^2(x^3) + x^-2 times 2sinmx^2 times cos^2?

Answer 7

it will be..
(-2x^-3) (sin^2(x^3)) + (x^-2) (2sin(x^3) cos(x^3) (3x^2)

Answer 8

got it now..??

Answer 9

can u explain the steps?

Answer 10

sure..

Answer 11

whenever you take derivative of trigonometric function then you have to take the derivative of its angle too...

Answer 12

which was the sin^2 rite?

Answer 13

like I did... you know derivative of sin is cos... but if the angle is not x then you have to take derivative of it too...
{(2sin(x^3) cos(x^3)} (3x^2)
now in curly brackets it is just derivative of sin^3(theta) and then (3x^2) is derivative of the angle (theta)...

Answer 14

so each is considered seperate and not together

Answer 15

yes..

Answer 16

but my professor gave this as the answer
6sin(x^3)cos(x^3)-2x^-3sin^2(x^3)

Answer 17

like derivative of sin^3(x)
it will be,
3sin^2(x) cos(x) as angle is simple "x" and derivative of x is 1 so we neglect it in this case.
now..
like derivative of sin^3(2x)
it will be,
3sin^2(2x) (cos(2x)) (2)

Answer 18

got it
but i dnt think we got the same answer as my professor though :/

Answer 19

yes.. by simplifying further you'll get the same answer..
(-2x^-3) (sin^2(x^3)) + (x^-2) (2sin(x^3) cos(x^3) (3x^2)
= (-2x^-3) (sin^2(x^3)) + (x^-2) (3x^2) 2 sin(x^3) cos(x^3)
= (-2x^-3) (sin^2(x^3)) + 6 (x^2)(x^-2) sin(x^3) cos(x^3)
as 3 is multiplied by 2 and (x^2)(x^-2) will be 1 so your answer will be
= (-2x^-3) (sin^2(x^3)) + 6 sin(x^3) cos(x^3)

Answer 20

@Dwade03 got it now..??

Answer 21

i got lost on your third step

Answer 22

(-2x^-3) (sin^2(x^3)) + 6 (x^2)(x^-2) sin(x^3) cos(x^3)

Answer 23

ok just wait a minute... i'll explain further...

Answer 24

thx

Answer 25

sir is it because first you multiply x^-2 to 2sinx^3 which gives 2sinx. then we multiply that by 3x^2 which results in 6sin(x^3)?

Answer 26

@Dwade03 got it now..??

Answer 27

yes but one doubt why didnt we multiply the 3x^2 to the cosx^3?

Answer 28

look if you multiply 3x^2 with cosx^3 you'll get..
3x^2 cosx^3 nothing more than this..

Answer 29

(x^3) infront of cos is its angle... you can not multiply angles with variables

Answer 30

oh ok.

Answer 31

like if you have 2cosx you can not write it like cos2x...

Answer 32

gotcha. thk u so much... are you like a tutor or something. just curious

Answer 33

yes I am lecturer in university... :)

Answer 34

oh wow... i wish i had you for my class. my professor is very confusing and doesn't explain things at all.

Answer 35

thanks.. but you'll have to bear your professor...

Answer 36

haha yes i know...i tried to get from tutors too but they didnt help much..either

Answer 37

don't worry i'll help you whenever you need me..

Answer 38

quite a tricky one this...

Answer 39

@cwrw238 ????

Answer 40

how can you find this y=4x\[\sqrt{x+\sqrt{x}}\]

Answer 41

all of them are together. i did it with a tutor but it was so confusing even though we got the answer

Answer 42

- i found it tricky thats all

Answer 43

your question is not clear..

Answer 44

\[y=4x \sqrt{x+\sqrt{x}}\]

Answer 45

now again 2 functions one is 4x and other is the rest one...
again product rule...

Answer 46

ok so 4x times the other one?