Question

Is the following true or false?

Answer 1

what is your question?

Answer 2

\[d^2/dx^2 (x^2\cos(x))=2\cos(x)-4x \sin(x)-x^2 \cos(x)\]

Answer 3

Differentiate the left side, or integrate the right:)

Answer 4

I would prefer the first

Answer 5

cant you do that on a scientific calculator?

Answer 6

Why?

Answer 7

The product rule is not as difficult, is it?

Answer 8

What is the derivative of:
1. \(\color{#000000}{\displaystyle \cos(x) }\)
2. \(\color{#000000}{\displaystyle x^2 }\)

can you tell me?

Answer 9

1. -(sinx)x'
2. 2x

Answer 10

Why x' ?

Answer 11

Is x a function of time, or a function of another variable?

Answer 12

x is the independent (or the "free") variable.

Answer 13

Your notation d/dx indicates that you are differentiating with respect to x. And the d²/dx² indicates the second derivative with respect to x.

Answer 14

(A different thing would be if you have dx/dt or something of this sort.... hope I clarified why you don't need x')

Answer 15

ok

Answer 16

For other than that,
-sin x
2x

are correct

Answer 17

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=\frac{d}{dx}\left[\frac{d}{dx}(x^2\cos x)\right] }\)

So, you need to apply the product rule inside the brackets... (hope I am not being abstruse)

product rule
\(\color{#000000}{\displaystyle \frac{d}{dx}(a(x)\times b(x))=a(x)b'(x)+a'(x)b(x) }\)

Answer 18

So, in your case,

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=\frac{d}{dx}\left[\frac{d}{dx}(x^2\cos x)\right] }\)

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=\frac{d}{dx}\left[2x\cos x-x^2\sin x \right] }\)

Answer 19

is this is understandable?

Answer 20

(if not, it's fine)

Answer 21

yea

Answer 22

Ok, now, what do you have to do?

Answer 23

find the derivative again?

Answer 24

yes

Answer 25

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=\frac{d}{dx}\left[2x\cos x-x^2\sin x \right] }\)

And you need to apply the product rule to difference components this time

Answer 26

So, you will need to find the derivative of:
1. 2x
2. cos(x)

Answer 27

1. 2
2. -sin(x)

Answer 28

Yes, correct

Answer 29

now, write the product rule

Answer 30

f`•g + f•g`

Answer 31

you know that
f=2x g=cos(x)
f`=2 g`=-sin(x)

Answer 32

\[2*\cos(x)+2x*(-\sin(x))\] right?

Answer 33

Yes, good (you can use \times or \cdot commands if you like)

Answer 34

Now we need to differentiate -x²sin(x) by the product rule.

Answer 35

(Doesn't matter which function is g and which one is f.)
But, if you set

f = -x² g=sin(x)

then,
f' = ? g' = ? (you tell me)

Answer 36

f'=-2x g'=cos(x)

Answer 37

Yes, awesome:

Answer 38

Now, write the entire product rule for this please

Answer 39

the two of them together?

Answer 40

or would that be skipping steps?

Answer 41

If you can write the two of them together, than please do.

Answer 42

(I always skip steps; I think it's fine)

Answer 43

\[2∗\cos(x)+2x∗(−\sin(x)) - 2x*\sin(x)-x^2*\cos(x)\]

Answer 44

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=\frac{d}{dx}\left[2x\cos x-x^2\sin x \right] }\)

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=\left[2\cos x-2x\sin x-2x\sin x-x^2\cos x \right] }\)

Yes, correct.

Answer 45

So, you had:

\(\color{#000000}{\displaystyle \frac{d^2}{dx^2}(x^2\cos x)=2\cos(x)−4x\sin(x)−x^2\cos(x) }\)

And after differentiating the left side twice we obtained

\(\color{#000000}{\displaystyle 2\cos x-2x\sin x-2x\sin x-x^2\cos x =2\cos(x)−4x\sin(x)−x^2\cos(x) }\)

Answer 46

you can go from there:)

Answer 47

thank you so much!!

Answer 48

yw