Question

find an equation of the tangent to the curve at the given point y=sin^2x (pi/6, 1)

Answer 1

take the derivative, substitute in your x and y values. that will give you the slope of the line. then find the y value of your point by substituting pi/6 into x in the original equation. you should then have the slope and a point. use point slope formula and determine your tangent line.

Answer 2

like I said, 1st find the derivative of your equation

Answer 3

I tried using the product rule and the chain rule, but its not working.

Answer 4

yeah Im trying but im not doing it right

Answer 5

You want to find the derivative of \[y=\sin^{2x}\], correct?

Answer 6

which is cos^2x

Answer 7

Ok, so what is the meaning of a derivative?

Answer 8

and equation of the tangent line

Answer 9

nope, the derivative is the slope of the tangent line

Answer 10

ok so now what? I just plug in pi/6?

Answer 11

Yes, plug pi/6 in to the derivative as x, the result will be the slope of the tangent line

Answer 12

but what about 1?

Answer 13

so I have 4cos^2(pi/6)?

Answer 14

\[\large y=\sin^2x=(sinx)^2\]\[\large y'=2(sinx)(sinx)'=2(sinx)(cosx)\]
Did you find the derivative ok?

Answer 15

When we take derivatives we work from the outside in.
We take the derivative of the OUTERMOST function, then apply the chain rule, multiplying by the derivative of the inside.

The problem with trig functions is the notation. Sometimes it's tough to recognize what the outermost function is because of the placement of the exponent.

Answer 16

so 4sin^2x becomes 4sinx cosx ?

Answer 17

Don't forget that factor of 2 in front!
4sin^2x would become 4[2(sinx)(cosx)]=8sinxcosx

Answer 18

Oh my goodness, zepdrix is correct. I did not check your derivation.

Answer 19

I apologize for that.

Answer 20

Confused lad? D':

Answer 21

so i plug pi/6 and 1 in seperately?

Answer 22

okay I get the chain rule. I just didnt see it when I looked at the problem.
so I have the derivative now what?

Answer 23

\[\large \sin^2(\frac{ \pi }{ 6 })\neq 1\]
Hmm I'm not sure where that (pi/6, 1) is coming from. It's not an ordered pair.
Oh oh oh oh. Is your problem suppose to be this?
\[\large y=4\sin^2x\]

Answer 24

Remember what a tangent line looks like?
It touches our curve at a particular point, and has the same slope as that curve at the given point. Our job is to find an equation for that straight line. We'll put it in the form y=mx+b.

Answer 25

yeah crap sorry its y=4sin^2x