Question

Find an equation of the tangent line to the curve
y=sin(3x)+cos(4x)
at the point (pi/6, y(pi/6))

Tangent line: y= ___

SHOW WORK!

I keep getting the wrong answer..
y'= 3cos(3x)-4sin(4x)
what do I do now?

Answer 1

Did you try to plug pi/6 in?

Answer 2

ur derivative is correct.... now plug in pi/6 into your derivative to get the slope of the tangent line at x=pi/6

Answer 3

yup, and I got it wrong.

Answer 4

pluggin in pi/6 gives me -2sqrt(3) is that right?

Answer 5

correct.... ^^^^ that's your slope

Answer 6

i'll b quiet...

Answer 7

I know.. and then I plug it into (y-y1)=m(x-x1)? right and plug in x and y

Answer 8

\[\begin{align}

y'= 3\cos(3x)-4\sin(4x)\\
y'=3\cos((3)\frac{\pi}{6})-4\sin((4)(\frac{\pi}{6}))\\
y'=0-2\sqrt{3}=-2\sqrt{3}

\end{align}\]
The slope is correct.

Answer 9

then do I plug in slope to (y-y1)=m(x-x1) or y=mx+b

Answer 10

\(y=mx+b\) of course!

Answer 11

I'll give you a hint, at which point do you want to find the equation of the tangent line? And what is this point's coordinates?

Answer 12

pi/6?

Answer 13

A coordinate should be in this format: \((x,y)\)

Answer 14

....?