Question

the y intercept of the line tangent to y=x sin x at x=pi is...
i don't understand this question :/

Answer 1

what course are you taking ?

Answer 2

calculus

Answer 3

ok, the slope of the line tangent at that point of the graph, will he y ' (pi)

First , find \[\frac{ dy }{ dx }=\frac{ d }{ dx }[x*\sin(x)]\]

Answer 4

ok the derivative is x cos x sinx

Answer 5

+ sinx i mean

Answer 6

The product rule, first times the derivative of the second, plus, the second times the derivative of the first.

\[\frac{ dy }{ dx } = x *\frac{ d }{ dx }\sin(x) + \frac{ d }{ dx }[x] * \sin(x)\]

Answer 7

\[\frac{ dy }{ dx } = x*\cos(x) + (1) *\sin(x)\]

Answer 8

yeah i understand that part :)

Answer 9

What is \[\frac{ dy }{ dx }\]when x=pi

Answer 10

uhm -pi ?

Answer 11

The slope of the tangent line at x=pi will be
\[\pi*\cos(\pi) + \sin(\pi)\]

Answer 12

yes which is -pi right

Answer 13

right

Answer 14

the tangent line at x=pi will be
\[y(x) = x*\sin(x)\]

\[y - y(\pi) = y~'~(\pi)*(x-\pi)\]

Answer 15

y(pi) = 0

y - 0 = -pi*(x-pi)

Answer 16

\[y = -\pi*x + \pi^2\]

Answer 17

that is the equation for the line tangent to y(x) at x=pi

Answer 18

ok so the y intercept is -pi?

Answer 19

The y-intercept can be found, by setting x=0
\[y=\pi^2\] ; x=0

the point

Answer 20

y-intercept
\[(0,\pi^2)\]

Answer 21

okay so how did u get pi^2?

Answer 22

one sec i have the graph...

Answer 23

using th point slope form for a line...

y-y1 = m*(x-x1)

the slope m is the derivative evaluated at x=pi ...m=-pi
x1 = pi
y1 = y(pi) = 0

the equation for the tangent line at x=pi is

y = -pi*x + pi^2

Answer 24

When x=0, y=pi^2

that is the y intercept of the line

Answer 25

pi^2 is about 9.87

Answer 26

wait i thought the y intercept was pi^2

Answer 27

it is, the graph i made has it in a decimal,, pi^2 = about 9387

Answer 28

9.87

Answer 29

ohhh okay that makes sense

Answer 30

The tangent line to a graph y(x) at x=a , is found by
\[y - y(a) = y~'(a)*(x-a)\]

Answer 31

just the point slope formula from algebra, with the slope being the derivative

Answer 32

point (a , y(a))
slope y '(a)

Answer 33

okay thanks. I think i get it the only thing I'm a little confused about is exactly how to get the y intercept..Like what were the steps u took to get pi^2

Answer 34

If you are given a function f(x). To get the equation for a line tangent at some point x=a to the graph. you use the point slope formula from algebra.

point (a , f(a))
slope f ' (a)
------------
point - slope formula for a line

y - f(a) = f ' (a) *(x - a)

Answer 35

Once you have the equation for the line tangent at x=a. you let x=0, to get the y-intercept

Answer 36

(0 ,y) - y intercept

Answer 37

The equation for this line was:
point (pi , y(pi)) = (pi , 0)
slope = y '(pi) = - pi
---------------------

Line point-slope form
\[y - 0 = -\pi*(x-\pi)\]\[y = -\pi*x + \pi^2\]

Answer 38

When x = 0, y = pi^2
\[y = -\pi*0 +\pi^2\]

the point \[(0,\pi^2)\]is the y-intercept of the tangent line to the graph at x=pi

Answer 39

ohh
okayy i get it

Answer 40

cool, i have time for another practice prob if you want to do another

Answer 41

okay thanks. uhm another one i have is : for what values of x is f(x)=|x+1| differentiable?
a) for all x can't equal 0, b) for all x is greater than or equal to 0
c) for all real numbers
d)for all x can't equal 1
e)for all x can't equal -1

Answer 42

A graph is not differentiable where there is a sharp point.

Answer 43

That is the graph of the absolute value of x, shifted to the left 1 unit

Answer 44

oh okay so x can't equal -1 ?

Answer 45

right, it is not differentiable when x=-1

there is no specific way to draw a tangent line at that point

Answer 46

lol, cant do it

Answer 47

okay yeah i see what u mean. thank u so much ! i really appreciate it :)