Question

Find an equation of the tangent line to the curve y=2sinx+sin2x at the point (0,0).

Answer 1

I can't figure this one out. Can someone talk me through it?

Answer 2

You need to find the derivative first

Answer 3

d/dx = -2 sin(x) - 2 sin(x) cos(x) sin(x)

Answer 4

\(\large\color{black}{ \displaystyle f(x)=2\sin x+\sin 2x }\)

\(\large\color{black}{ \displaystyle f'(x)=2\cos x+2\cos 2x }\)

Answer 5

do you see how I got this?

Answer 6

Yeah. For some reason I was using a different equation.

Answer 7

d/dx [ 2sin(x) ] = 2 cos(x)

d/dx [ sin(2x) ] = cos(2x) × {d/dx 2x] = cos(2x) × 2 = 2cos(2x)

Answer 8

The derivative of the function is the function's slope.
And thus, the derivative evaluated at x=0 (i.e f'(0)) is the slope of the function at x=0.

Answer 9

The 2 in the second part is supposed to be to the second power.

Answer 10

\(\large\color{black}{ \displaystyle f(x)=2\sin x+\sin^2x }\)

this way?

Answer 11

Yes

Answer 12

\[2\cos(x)+2\sin(x)\cos(x)\]

Answer 13

\(\large\color{black}{ \displaystyle f(x)=2\sin x+(\sin x)^2 }\)

\(\large\color{black}{ \displaystyle f'(x)=2\cos x+2(\sin x)^{2-1} \times \cos x}\)

Answer 14

see what I did?

Answer 15

Yep. So, after that you plug in 0 for x which makes the slope 2. Correct?

Answer 16

THat's right

Answer 17

So you got:

slope = 2
point: (0,0)

do your line now:)

Answer 18

Awesome. I was just using the wrong equation. Thanks guys!

Answer 19

Anytime:)