Question

Find an equation of the tangent line to the curve y=2x sin x at the point (pi/2,pi)

Answer 1

first, find the slope (m) = y'

Answer 2

find dy/dx at x=pi/2

Answer 3

Wouldn't that be 2cos x? and if you plug in pi/2 or 90 degrees you get 0?

Answer 4

use the product rule
y ' = vu' + uv"

Answer 5

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Answer 6

then eq. of tangent is
\[y-\pi=(\frac{ dy }{dx }at x=\frac{ \pi }{2 })\left( x-\frac{ \pi }{2 } \right)\]

Answer 7

Okay after I do the product rule, which I end up with 2, what step do I take from there?

Answer 8

And after plugging in x*

Answer 9

yes, correct m = 2

then use the equation like @surjithayer said above :

Answer 10

Phew thank you! While Im here can you I ask of you for help on another problem?

Answer 11

what's the question then ..

Answer 12

Differentiate. f(x)= xe^x csc x

Answer 13

let u = xe^x and v = cscx
use the product rule

Answer 14

to get u' looks we need the product rule again :)

Answer 15

u = xe^x
u ' = (xe^x)' = x' e^x + x (e^x)' = 1 e^x + x e^x = e^x + xe^x, agree ?

Answer 16

Oooh okay, haha yeah I was just confusing myself I agree yes, Thank you for your help!

Answer 17

oke, go back the original question :)
f(x)= xe^x csc x
f '(x) = (xe^x)'csc x + xe^x(csc x)'
(xe^x)' already done above, now (cscx)' = -cscxcotx
so, f ' = u'v + uv' = ....

Answer 18

xe^xcsc + e^x cscx-xe^x cscx-xe^xcscx + xe^x cotx, and I can see that you can just factor out an xe^xcsc so final answer would be.. e^xcscx(-x cot x +x +1) ?