Question

find the derivative of x^3 * cosX ...would like to compare my steps with yours please

Answer 1

use the product rule./

Answer 2

used product rule first

Answer 3

okay, and what were you getting?

Answer 4

3x^2(cosx) + x^3(-sinX)

Answer 5

yes that is correct. you can re-write it a bit better though.

Answer 6

factor x^2?

Answer 7

well,
\(\large\color{black}{ 3x^2(\cos x) + x^3(-\sin x) }\)
is same as,
\(\large\color{black}{ 3x^2 \cos x - x^3 \sin x }\)

Answer 8

is XsinX an identity?

Answer 9

you don't need to factor. If that is what your teacher would required or (even) prefer, then you can do that, that is not a problem:)

Answer 10

I don't think there is really an identity for \(\large\color{black}{ x\sin x }\).

Answer 11

x^2(3cosX - XsinX)

Answer 12

yes:) you can do that too, I don't see a need though, unless you are going to be finding the critical numbers.

Answer 13

what would the equation for the tangent line at x=pi be?

Answer 14

you will need to plug in pi, into the derivative. this will give you the slope.
then the point is

(pi, f(pi) )

Answer 15

\(\large\color{black}{ f(x)=x^3\cos x}\)

\(\large\color{black}{ f(\pi)=(\pi)^3\cos (\pi)}\)

\(\large\color{black}{ f(\pi)=(\pi)^3(-1)}\)

\(\large\color{black}{ f(\pi)=-(\pi)^3}\)

Answer 16

so your point is, \(\large\color{black}{ (\pi,~-(\pi)^3~)}\)

Answer 17

then \(\large\color{black}{ f \prime(\pi)}\) is the slope.

Answer 18

We had: \(\large\color{black}{ f\prime(x)=x^2(3 \cos x - x \sin x)}\)

Answer 19

for the answer i get

Answer 20

can you find, \(\large\color{black}{ f\prime(\pi)}\) ?

Answer 21

the equation at pi is 3(pi)^2x - 4(pi)^3

Answer 22

I found the point for you, and you need to find the slope.
then apply the point slope formula.

Answer 23

3((pi)^2)(x) - 4(pi)^3

Answer 24

I don't understand what yu aregetting, and after plugging pi for x, why do you still have an x?

Answer 25

equation of the tangent line at that point

Answer 26

equation has " = "

Answer 27

\(\large\color{black}{ f\prime(\pi)=(\pi)^2(3 \cos\pi- \pi \sin \pi)}\)
\(\large\color{black}{ f\prime(\pi)=(\pi)^2(3 \cos\pi- \pi \sin \pi)}\)
\(\large\color{black}{ f\prime(\pi)=(\pi)^2(3 \cos\pi)}\)
\(\large\color{black}{ f\prime(\pi)=(\pi)^2(3)(-1)}\)

\(\large\color{black}{ f\prime(\pi)=-3(\pi)^2}\)

Answer 28

\(\large\color{black}{ f\prime(\pi)=m=-3(\pi)^2}\)

Answer 29

\(\large\color{black}{ (y-y_1)=m(x-x_1)}\)

Answer 30

actually got -3(pi^2)x + 2(pi^3)

Answer 31

got for what?

Answer 32

the slope of the tangent line

Answer 33

y = -3(pi^2)x + 2(pi^3), sorry, the equation of the tangent line

Answer 34

slope: \(\large\color{black}{ f\prime(\pi)=-3(\pi)^2 }\)
point: \(\large\color{black}{ (\pi,-\pi^3) }\)

Answer 35

\(\large\color{black}{ y-(-\pi^3)=-3\pi^2~(x-\pi) }\)
\(\large\color{black}{ y-(-\pi^3)=-3\pi^2x+3\pi^3 }\)
\(\large\color{black}{ y+\pi^3=-3\pi^2x+3\pi^3 }\)
\(\large\color{black}{ y=-3\pi^2x+2\pi^3 }\)

Answer 36

you were correct, I was just checking that.

Answer 37

awesome, for some reason, wolfram alpha spits out a different answer for our derivative, but it looks like we did it correctly, maybe it's their algorithm?

Answer 38

you think we're right, though?