Question

LGBADERIVATIVE:
\[\huge y = x^2 \sin x \tan x\]

product rule??

Answer 1

do i change it to \[\huge y = x^2 (\frac{\sin^2 x}{\cos x})\]
then use product rule?

Answer 2

Ya i would try then lgba

Answer 3

So we will let f (x) = x^2sin(x)tan(x) so the derivative will require the product rule!

Answer 4

But usually when we do a product rule we are only working with 2 expressions. Here we have three though.

So this rule looks like for\[ \huge\ f(x) = a*b*c \]where a, b, and c are all functions of x, we have:

\[\huge\ f '(x) ab*d(c)/dx + a*d(b)/dx*c + d(a)/dx*bc\]

Answer 5

I am going correctt @lgbasallote

Answer 6

so how to product rule this...

Answer 7

i think u r going right @Rohangrr

Answer 8

what is the derivative of \[\frac{\sin^2 x}{\cos x}\]

Answer 9

So we have\[ f '(x) = x^2\sin(x)*d(tanx)/dx + x^2*d(sinx)/dx*\tan(x) + d(x^2)/dx*\sin(x)\tan(x)\]

So \[f '(x) = x^2\sin(x)\sec^2(x) + x^2\cos(x)\tan(x) + 2xsin(x)\tan(x) \]

So \[f '(x) = x^2\sin(x)\sec^2(x) + x^2\sin(x) + 2xsin(x)\tan(x) \]
final answer
\[\huge\ =\sin(x)[x^2\sec^2(x) + x^2 + 2xtan(x)]\]

Answer 10

lol wow nice

Answer 11

it looks great @Rohangrr

Answer 12

medals

Answer 13

let's see if it's right...

Answer 14

lol

Answer 15

if some one blocks then how to give medal ?

Answer 16

undertaker

Answer 17

there is a easier mwthod deffrentiate both wid respect to dx

Answer 18

\[\huge \frac{x^2}{\sin x} + x^2 \sin x + \frac{2x \sin^2 x}{\cos x}\]

Answer 19

lol wow...this is a hard integral...

Answer 20

k so lgba i am correct

Answer 21

like i said..im going to check first

Answer 22

yes u r right @lgbasallote first check :D

Answer 23

heh this is hard interal...im just gonna derive..

Answer 24

best of luck my friend :)

Answer 25

thanks saif u forgot bout the mail

Answer 26

@Rohangrr , OHHHH! Man! My bad. I totally forgot. Please forgive me! :'(
I will do it soon! :D

Answer 27

yup @Rohangrr made a mistake

Answer 28

lol

Answer 29

np

Answer 30

^_^

Answer 31

pls check guys...here's what i got

\[\huge \sin x (2x^2 + 3x\tan^2 x)\]
am i right or @Rohangrr ?

Answer 32

what do u mean @LGBA
"am i right or @Rohangrr""

Answer 33

we have different answers...so who's right

Answer 34

lol @saifoo.khan

Answer 35

idk the answer too im just checking my work

Answer 36

;)

Answer 37

saif is ........ hmmmm........ why were u worrying so mch @lgbasallote

Answer 38

worrying?

Answer 39

chillllll okay okay leave it

Answer 40

lol so which is correct?

Answer 41

im trying to see if they're actually equal...

Answer 42

oops i made a mistake...but it's still not similar to @Rohangrr 's answer...

\[\huge \sin x (2x^2 + \tan^2 x + 2x\tan x)\]

Answer 43

wait in that some part is hiding i ll do it

Answer 44

waitttt why not change i to..
\[y = x^2sinxtanx => x^2*\frac{1-\cos^{2}x}{cosx} => x^2*\frac{1}{cosx} - \frac{\cos^{2}x}{cosx} => x^2secx-cosx\]

Answer 45

\[y = x^2sinxtanx => x^2*\frac{1-\cos^{2}x}{cosx} => x^2*\frac{1}{cosx} - \frac{\cos^{2}x}{cosx} =\]
\[ => x^2secx-cosx\]

Answer 46

my 3rd conversation
where a, b, and c are all functions of x, we have:
\[f '(x) ab*d(c)/dx + a*d(b)/dx*c + d(a)/dx*bc \]

Answer 47

what is that? lol

Answer 48

btw, i want to say something off topic; am i the only who is experiencing problems with this site???

Answer 49

nope

Answer 50

so now we have 3 different answ

Answer 51

well other way is:
\[\huge\ ycos(x) = x^2\sin^2(x) \]

Using implicit differentiation treating y as a function of x we have:

\[y(-sinx) + \cos(x)dy/dx = x^2[2\sin(x)\cos(x)] + 2xsin^2(x) \]

So isolating dy/dx on the left side of the equation:

\[\cos(x)dy/dx = x^2\sin(2x) + 2xsin^2(x) \]

So \[\huge dy/dx = [x^2\sin(2x) + 2xsin^2(x)] / \cos(x)\]

Answer 52

noooo..im suggesting you to change it to
\[ x^2secx-cosx\]
and apply the product rule; it wont look messy

why implicit differentiation????

Answer 53

yup it turnsout as a completely new answer4

Answer 54

that 4 does not mean anything

Answer 55

lol u all are getting confused and confusing me

Answer 56

so who's right??

Answer 57

I think question by answer not medal

Answer 58

lol what?

Answer 59

wait. is this for a bigger problem?

Answer 60

i believe so..

Answer 61

QUESTION --> answer
QUESTION -\-> medal

Answer 62

i still dont get what you mean

Answer 63

hmm i giuess no one can do it haha

Answer 64

what is the problem? my method would work; however, tell us what's the problem that you're working on?

Answer 65

lol look at the blue box :p

Answer 66

lol i know..the bigger problem towards the question? that question must have been derived from something

Answer 67

the book :P lol