Question

differentiate using quotient rule y=tanx

Answer 1

\[
y=\tan x=\frac{\sin x}{\cos x}\\
\mathbf{Quotient\ rule:}\left(\frac f g\right)'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}\\
f(x)=\sin x,\ g(x)=\cos x
\]

Answer 2

Can you take it from there?

Answer 3

no whats the next steps pls

Answer 4

Next step is to find out what f'(x) and g'(x) are, and then just plug it all into the formula for the quotient rule.

Answer 5

pls can u tell me what f'(x) and g'(x)are?

Answer 6

The derivative of sin x is cos x, the derivative of cos x is -sin x. You must memorize these, they come up very often.

Answer 7

k

Answer 8

Can you now solve for the derivative of tan x?

Answer 9

nt really s the nxt thing gng to be f'(x)=cosx and g'(x)=-sinx

Answer 10

That is correct. Now just plug everything into the quotient rule.

Answer 11

k that will be cosx(cosx)-sinx(-sinx)/cos^2x
am i ryt?

Answer 12

Can you simplify that further?

Answer 13

no hw s it done pls?

Answer 14

Well first just rewrite it nicely, \[\frac{\cos^2x+\sin^2x}{\cos^2x}\] Do you recognize that numerator?

Answer 15

no wat s it

Answer 16

It's the most fundamental trig identity that there is. You must be able to recognize what \(\sin^2x+\cos^2x\) is.

Answer 17

k wat are they

Answer 18

http://www.youtube.com/watch?v=91Jxfu4FntM

Answer 19

tnx that hlped so it is = 1 ryt?

Answer 20

nbouscal is the answer gng to be 1 nd am done?

Answer 21

The numerator simplifies to 1. So you have \(\dfrac{1}{\cos^2x}\). That can be written a simpler way also.

Answer 22

so what is the final ans?

Answer 23

What is another way of writing \(\dfrac{1}{\cos x}\)?

Answer 24

1/-sinx ?

Answer 25

No. This is basic stuff that you should have learned long before learning anything about differentiation. You should review your trigonometry, maybe using some practice exercises on Khan Academy.

Answer 26

k wat is it

Answer 27

I'm not here to give you answers. This is something you can find out for yourself very, very easily.

Answer 28

k thnks