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Differentiate the following with respect to x

sin^2 (x) / cos^3 (x)


Erm... :o:
Reply 1
Avatar for Pheylan
Pheylan
have you come across the quotient rule?

sin2xcos3x\frac{sin^2x}{cos^3x}

let u=sin2xu=sin^2x and v=cos3xv=cos^3x
Quotient rule? Or write as sin⁡2xcos2x⋅secx=tan2xsecx\frac{\sin^2x}{cos^2x}\cdot secx=tan^2xsecx and then either use an identity connecting tan and sec or use the product rule.
Reply 3
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
ye i've tried to use thazt method. but i get:

2cos^4 (x) sin (x) + 3cos^2 (x) sin^3 (x)
____________________________________________

cos^6 (X)
Quotient rule

(d/dx)f(x)/g(x)=
[f'(x)g(x)-g'(x)f(x)]/(g(x)^2)

Sorry but I suck with LaTex, so if that just looks a mess, look on wikipedia or something for the quotient rule, it's right what they've got written for it.
Reply 5
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
i fully understand the quotient rule..tbh is my answer correct? just not simplified enough??
Reply 6
Avatar for DFranklin
DFranklin
18
FoOtYdUdE
ye i've tried to use thazt method. but i get:

2cos^4 (x) sin (x) + 3cos^2 (x) sin^3 (x)
____________________________________________

cos^4 (X)
(a) Check your denominatoer.
(b) Cancel out what you can, then take a common factor of the top and remember sin^2+cos^2 =1.
Reply 7
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
answer in textbook =

sinx(2+sin^2 x)
_______________
cos^4 x
Reply 8
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
soz, it was cos^6 x not cos^4 x (denominator)
Reply 9
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
DFranklin
(a) Check your denominatoer.
(b) Cancel out what you can, then take a common factor of the top and remember sin^2+cos^2 =1.

erm..:o:
FoOtYdUdE
erm..:o:
As uninformative posts go, this scores pretty highly.

What is the problem?
Reply 11
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
DFranklin
As uninformative posts go, this scores pretty highly.

What is the problem?

Sorry \about this. Hang on.
Reply 12
Avatar for FoOtYdUdE
FoOtYdUdE
OP
9
This is my working:

Is it coreect? Have I gone wrong anywhere??


(2cosxsinx.cos^3 x - (-3cos^3 x sinx . sin^2 x) ) / ( cos^3 x)^2 )

= (2 cos^4 x sinx + 3cos^2 x sin^3 x ) / (cos^6 x)
Reply 13
Avatar for Rubgish
Rubgish
13
FoOtYdUdE
This is my working:

Is it coreect? Have I gone wrong anywhere??


(2cosxsinx.cos^3 x - (-3cos^3 x sinx . sin^2 x) ) / ( cos^3 x)^2 )

= (2 cos^4 x sinx + 3cos^2 x sin^3 x ) / (cos^6 x)


Thats fine, just cancel down. Take out common factors, and remember that sin^2 x + cos^2 x = 1
As has already been said above, you should really try and simplify expressions before you differentiate them.

Read up to see how this expression goes to tan^2xsecx
Which can be found using the product rule.

Or it cancels even further to (sec^2x - 1)secx = sec^3x - secx
Which can be found using the chain rule.

So, would you rather use the chain rule, the product rule, or the quotient rule?

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