Question

𝑓(𝑥) = 𝑥^2 + 1, 𝑔(𝑥) = √𝑥.
Determine the derivative of 𝑖) 𝑓(𝑔(𝑥)), 𝑖𝑖) 𝑔(𝑓(𝑥)),
𝑖𝑖𝑖) 𝑓(𝑓(𝑥)), and 𝑖𝑣) 𝑔(𝑔(𝑥)) with respect to 𝑥

Answer 1

@Vocaloid

Answer 2

Just need a better explanation

Answer 3

well

i) is just asking for the derivative of 𝑓(𝑔(𝑥)), right? so you just take g(x) = sqrt(x) and plug that into the x-value for f(x), then take the derivative of the result

Answer 4

would the exponent 2 go in the sqrt too or no

Answer 5

No it'd go outside the square root.

f(x) = x^2
g(x) =√x

f(g(x))
= f(√x)
= (√x)^2 + 1

get it?

Answer 6

okay is that it

Answer 7

No you gotta take the derivative of that. You know how to take the derivatives right?

Answer 8

um kind of can you just clear up the concept

Answer 9

Yeah so the first question asks you to take the derivative of f(g(x))

and you know f(g(x)) is (√x)^2+1
which can be simplified to x + 1

When you take the derivative of something, you basically use this formula:
\[\frac{d}{dx}ax^b =(b*a)x^{b-1}\]

so x + 1 can be rewritten as
\[1x^{1}+1x^0\]

can you find the derivative of that?

Answer 10

x^-1

Answer 11

oh shoot wait I made a mistake. Sorry.

1x^1 + 1

Since 1 is a constant, the derivative of constants is always 0.

so the derivative is actually
\[(1*1)x^{1-1}+0\]
which is just 1.

Answer 12

I can give you some more examples of derivatives:

derivative of \[\frac{d}{dx}(5x^3+2x^2+x+4) = (3*5)x^2+(2*2)x+(1) = 15x^2+4x+1\]

Answer 13

idk if it'll make this more/less confusing but generally if you have something like

A^B

you bring the exponent down as a coefficient to get

B * A^B

then reduce the exponent by 1 to get

B* A^(B-1)

then multiply by the derivative of the base (A)

--> B * A^(B-1) * derivative of A

Answer 14

it just takes a bit of time to learn the pattern that's all

Answer 15

I'm taking notes on this so we can continue in like 5 mins

Answer 16

ok

Answer 17

okay so derivative of sqrtx^2+1 is 2x^2?

Answer 18

Well actually

sqrtx^2 + 1 = x + 1 right?

Answer 19

so derivative of x + 1 is ?

Answer 20

It's just 1. Derivative of x + 1 is just 1.

Derivative of x is 1.
Derivative of 1 (or any constant) is 0.
So it becomes 1 + 0.

Do you understand why?

Answer 21

Yes I do understand why

Answer 22

Could you just explain what happens to the exponent though

Answer 23

Yeah so the exponent gets brought down in front of x as a multiplier, and the exponent on top decreases by 1.

x^3 = 3x^2
x^2 = 2x
4x^2 = (4*2)x = 8x

Answer 24

No I mean in the sqrtx^2+1 problem

Answer 25

taking the square root of something will undo a square

Answer 26

sqrt(A^2) = A

Answer 27

Got it

Answer 28

oh yeah.

you can also think of it like this:
\[\sqrt{x} = x^{\frac{1}{2}}\]

and
\[(x^a)^{b} = x^{a*b}\]

so
\[\sqrt{x}^2 = (x^{\frac{1}{2}})^2 = x^{\frac{1}{2}*2} = x\]

Answer 29

anyway that's the first derivative done; derivative of 𝑖) 𝑓(𝑔(𝑥)) = 1 according to our calculations

if you want to start the second one 𝑖𝑖) 𝑔(𝑓(𝑥)) I'm ready when you are

Answer 30

OKay so
f(x)=x^2+1
g(x)=sqrtx
g(f(x))
g(sqrt^2)

Answer 31

Wait another question :S where did the +1 come from in the first one :/

Answer 32

it came from f(x) = x^2 + 1

Answer 33

oh right

Answer 34

OKay so
f(x)=x^2+1
g(x)=sqrtx
g(f(x))
g(sqrt^2)+1
=1

Answer 35

Well actually.

g(f(x)) is g( sqrt^2+1 )

Answer 36

I thought we were on ii

Answer 37

but yeah you need to pay attention to parentheses

Answer 38

so ii is g( sqrt^2+1 )

Answer 39

yes

Answer 40

now applying g(x) = sqrt(x)

g(f(x)) = g( sqrt^2+1 ) = sqrt(x^2+1) then take the derivative of this

Answer 41

x?

Answer 42

re-write sqrt(x^2+1) as (x^2 + 1)^(1/2) and apply the derivative rule

Answer 43

1/2(x^2+1)^-1/2

Answer 44

almost, you need to also multiply by the derivative of x^2 + 1

Answer 45

2x+1?

Answer 46

2x?

Answer 47

:S

Answer 48

x^2 + 1 ---> derivative is 2x (since 1 is just a constant)

so (x^2 + 1)^(1/2) derivative ---> (1/2)(2x)(x^2+1)^(-1/2) ---> x * (x^2+1)^(-1/2) = x/sqrt(x^2+1) = your derivative

Answer 49

zarkam have you learned the chain rule before?

Answer 50

\[\frac{ x }{ \sqrt{x^2+1} }\]

Answer 51

No , I haven't my professor is reallly not helping any of us which is why me along with other students are struggling

Answer 52

he goes over problems and says to solve them you use the chain rule. We ask him what the rule is and he says we haven't gotten there yet

Answer 53

okay for iii it would be

f(f(x)

f(x^2+1)

Answer 54

I think :S

Answer 55

Yeah you're right so far. me/voca will help you learn the chain rule

Answer 56

2x as the derivative

Answer 57

f(x^2+1)

since f(x) = x^2 + 1, then you need to plug "x^2+1" into f(x) to get

(x^2+1)^2 + 1

then take the derivative of that

Answer 58

2x^2?

Answer 59

orrr 2(2x)

Answer 60

I think Vocaloid wants you to simplify that before you take the derivative.

Just simplify (x^2+1)^2 + 1 first

Answer 61

(x^2)^2

Answer 62

Well actually
\[(x^2+1)^2 + 1 = (x^4 + 2x^2 + 1) + 1 = x^4 + 2x^2 + 2\]
right?

Answer 63

Yes

Answer 64

Alright so find the derivative of
\[x^4 + 2x^2 + 2\]

Answer 65

(4)x^3+(2x2)x+2=4x^3+2x^2

Answer 66

(4)x^3+(2x2)x+2=4x^3+6x^2

Answer 67

uh, check that last one

Answer 68

4x^3 + (2x2)x = 4x^3 + 4x

did you see what you did wrong?

Answer 69

Yess

Answer 70

okay well there's your answer

Answer 71

See how long it takes to do this? I wanna show you a trick:

If you have a function like
f(x) = x^2

and you have another function like
g(x) = x^3

and you want to find the derivative of

f(g(x))

you can rewrite it as

f'(g(x)) * g'(x)

so

f'(x) = 2x and f'(g(x)) = 2(g(x)) = 2(x^3)
g'(x) = 3x^2
f'(g(x))g'(x) = (2x^3)(3x^2)

does that make sense?

Answer 72

f'(x) means "the derivative of f(x)" btw

Answer 73

Yeah it makes more sense than what we went over before