Question

equation of the tangent line??

Answer 1

Have you found y' yet?

Answer 2

because m is actually equal to y' evaluated at x equals pi

Answer 3

is it 5(-sinx) ??

Answer 4

You will have to use product rule to find y'

Answer 5

you have 5x*cos(x)

Answer 6

\[y=5x \cdot \cos(x) \\ y'=(5x)'\cos(x)+5x(\cos(x))'\]
can you take it from here?

Answer 7

the derivative part that is

Answer 8

If \(f(x)\) is our function, then the equation of the tangent line to the function at the point \((a,f(a))\) is given by: \(y=f(a)+f^{\prime}(a)(x-a)\), where \(f^{\prime}(a)\) is the derivative of \(f(x)\) evaluate at \(x=a\).

In this problem, you're given that \(f(x)=5x\cos x\) and the point \((\pi, -5\pi)\). We now see that \(f^{\prime}(x) = (5x)^{\prime}\cos x + 5x(\cos x)^{\prime}=\ldots\) by the product rule. We furthemore see that \(f^{\prime}(\pi)=\ldots\).

Plugging all of this into the tangent line equation gives us \(y=f(\pi)+f^{\prime}(\pi)(x-\pi)=\ldots=mx+b\).

Once you simplify the tangent line equation, you'll have the values you're looking for.

Can you take things from here?

Answer 9

derivative is -5xsinx + 5cosx

Answer 10

looks great

Answer 11

now in chris's thing there
he mentions you need to take your derivative and evaluate it at pi to find m

Answer 12

so then you plug in pi to this equation -5sinx+5cosx

Answer 13

well you plug in the -5xsin(x)+5cos(x)

Answer 14

and the slope is -5

Answer 15

sounds good
so you have point slope form is
\[y-y_1=m(x-x_1) \\ y-y_1=-5(x-x_1) \text{ because you found the slope m=-5 }\]

Answer 16

now you know a point on the line (pi,-5pi)

Answer 17

i got 0 for the y-intercept

Answer 18

plug in your point where you have (x1,y1)

Answer 19

oh i just plugged it into y+mxtb

Answer 20

you mean the y=mx+b ?

Answer 21

that works too

Answer 22

and you said you got b is 0?

Answer 23

yea thats it

Answer 24

alright sounds good

Answer 25

thank you very much