Question

Why does this suddenly go away?

Differentiate the function y=(8x-7)^4
I am using the power rule
g(x)=8x-7
k=4

In the example, after some work it shows
dy/dx=4(8x-7)^3*d/dx(8x-7)
then
=4(8x-7)^3*8

What happened?

Answer 1

@Psymon

Answer 2

OH, it's the deriv of 8x-7

Answer 3

Yeah, chain rule. YOu take the derivative of whats inside, too.

Answer 4

So dy/dx=32(8x-7)^3

Answer 5

So now for a new equation: y=(3x+4)^5

Answer 6

g(x)=3x+4
k=5

Answer 7

dy/dx=5(3x+4)^5-1*d/dx(3x+4)

Answer 8

Still going to be chain rule. You can kind of think as all derivatives as chain rule. You will always take the derivative of each inner function. If you have 3 exponents or things that barrier off expressions, then you have 3 derivatives you have to compute. Its just when you first do derivatives, the chain rule doesnt show up because the derivative of aninner function might be one and not matter. But yeah, keep going, lol.

Answer 9

d/dx(3x+4)=3

Answer 10

dy/dx=5(3x+4)^4*3
15(3x+4)^4

Answer 11

Looks good tome.

Answer 12

Hmm okay, so for a square root problem I will simply change the square root over to a power?

Answer 13

Yeah, realize a sqrt is a (1/2) power, meaning you will be multiplying by 1/2 and lowering the power to a -1/2

Answer 14

y=sqrt(8-7x)
y=(8-7x)^1/2
k=0.5
g(x)=8-7x
0.5(8-7x)^-0.5*d/dx(8-7x)
d/dx=8
0.5*8=4
4(8-7x)^-0.5

Answer 15

Hmm well mymathlab says it's wrong..

Answer 16

\[\sqrt{8-7x} = \frac{ 1 }{ 2 }(8-7x)^{-1/2}(-7)\]
\[\frac{ -7 }{ 2\sqrt{8-7x} }\]

Answer 17

Ookay the example is way different :O? Weird.

Answer 18

Your answer is right, but how did you get -7 for d/dx?

Answer 19

derivative of 8-7x. Looks like you did it as if it were 8x-7 and not 8-7x.

Answer 20

Oh! Yes, I did. I was like how in the world did he get his constant into play?!

Answer 21

Lol, whoops xD

Answer 22

Okay:)
y=(2x^2-5)^-10
k=-10
g(x)=2x^2-5

-10(2x^2-5)^-11*d/dx(2x^2-5)
d/dx of 2x^2-5 drop the constant, 2x^2, bring down power, multiply, subtract power by 1 and you get 4x.
so now,
-10(2x^2-5)^-11*4x

Answer 23

Looks fine :3

Answer 24

Yikes! A factor!
y=1/(2x-1)^3

Answer 25

Just help with the first step and then I can do it.. :)

Answer 26

Not bad, theres a trick to simply fraction ones:
\[\frac{ -6 }{ (2x-1)^{4} }\]

Just if you can see how I got that xD First step would be make the whole denominator into a negative exponent and put it in the numerator.

Answer 27

wat

Answer 28

Here's the first step:

\[(2x-1)^{-3}\]

Answer 29

How did you get that?

Answer 30

Oh wait nevermind lo.

Answer 31

You can always change the sign of an exponent by moving it to the opposite side of a fraction.

Lol, okay, haha.

Answer 32

I thought the original equation was (2x-1)^4 for a second, I was like What in the world did he just do!?

Answer 33

Theres a trick to simple fractions where I can give you derivatives of them very fast.

Answer 34

I'd like to learn it after this problem:)

Answer 35

If it makes sense, sure, lol.

Answer 36

Its not really a shortcut as much as it is pattern recognition and knowing whats going to happen.

Answer 37

Okay so after (2x-1)^-3, you do what?

Answer 38

Just chain rule it up and solve?

Answer 39

Do what youve been doing. k = -3, it multiplies down, lower power by 1, chain rule it out, etc xD

Answer 40

f(x)=(7+x^5)^5-(5+x^7)^3

Answer 41

Chain rule by themselves then subtract in the end?

Answer 42

Nothing new here. Yeah, work with each term separately,

Answer 43

(7+x^5)^5
k=5
g(x)=7+x^5
deriv: 5x^4

5(7+x^5)^4*5x^4

k=3
g(x)=5+x^7
deriv 7x^6

3(5+x^7)^2*7x^6

5(7+x^5)^4*5x^4-3(5+x^7)^2*7x^6

Answer 44

Yeah, looking good :3

Answer 45

Now how would you simplify this? Do I literally have to expand it or..

Answer 46

\[25x^{4}(7+x^{5})^{4}- 21x^{6}(5+x^{7})^{2}\]Yeah, hard to do much with it.

Answer 47

Hmm how did you get 25?

Answer 48

I have the same thing you have, its just multiplied a bit.

Answer 49

How old are you? Jw

Answer 50

Eh. 23.

Answer 51

You're very smart haha.
f(x)=sqrt(x)+(x-3)^4

Answer 52

Sometimes yes, sometimes no, lol.

Answer 53

Same thing, just take the derivative of each termalone.

Answer 54

deriv of sqrt(x) is 1/2sqrt(x)

So moving onto the next.
4*(x-3)^3*1

1/2sqrt(x)+4(x-3)^3

Answer 55

Right. Assuming you mean for the sqrt to be in the denominator, then yes

Answer 56

I'm only seventeen:( Haha

Answer 57

Well, Ive had more time to learn more? xD

Answer 58

y=x^2sqrt(10x-3)

Answer 59

SO finally a product rule, eh.

Answer 60

I know :P I just like to have an excuse for being dumb.

Answer 61

Alright so looking at the example this does not look that hard!
f(x)=x^2
g(x)=sqrt(10x-3)

Answer 62

Haha, thats nice to have xD

Answer 63

Yeah, its just extra steps. Easy stuff, though.

Answer 64

Do you have a degree in math or a math-related field?
And can you walk me through this? The problem has an error after the third step and won't load

Answer 65

Im working on a math degree, but im still a student xD Um...yeah,
f = x^2
g = sqrt(10x-3)

So derivative of f times regular g + derivative of g times regular f. A lot of text books show the reverse order, but both work. As far as Im concerned, f(x) is first, so Im doing f'(x) first, haha. SO yeah, full formula is f'(x)g(x) +f(x)g'(x) So doing that I get:

\[2x \sqrt{10x-3}+x^{2}(10)(\frac{ 1 }{ 2 })(10x-3)^{-1/2}\]

Answer 66

That answer haha, when I was typing it in I was like "No way is this right, it's all over the place"
Guess not ;_;

Answer 67

Nah, some of these get ridiculous, its normal xD

Answer 68

f(x)=((x-6)/(x+3))^5

Answer 69

Everything to the 5th, correct?

Answer 70

So yeah, it's a fraction of x-6 over x+3, all to the power of 5

Answer 71

Alrighty. Nothing hard, just requires you to know what youre doing and not doing, just as importantly.

Answer 72

So k=5
g(x)=x-6/x+3

Answer 73

Oh god please tell me I don't have to use the quotient rule.

Answer 74

Not really. Here is how I like to tell people to think of it:



f is still going to be x-6, g being x -3. ANd you donthave to use quotient rule, you can do product rule.