Question

SINGLE-VAR CALCULUS

whole problem: "Find an equation to the tangent line to the curve at the given point"
with given equation -
y=log2(x^2-3x+6) at (2,2)

Step one - derivative of:\[y=\log{2}(x^{2}-3x+6)\]
I thought I'd use the Product Rule but Symbolab tells me to leave the \(\log{2}\) alone. :\

Answer 1

why would you lead him on, WHY WOULD YOU LEAD THIS WHOLE THING ON?

Answer 2

What is the question?

Answer 3

@Zale101

Answer 4

\[y=\log_2(x^2-3x+6)\]???

Answer 5

log(2) not log-base 2

Answer 6

\[y=\log2(x^2-3x+6)\]

Answer 7

Yes, that.

Answer 8

Here is the rule.

Answer 9

Could you apply the rule of differentiation ?

Answer 10

What do you mean?

Answer 11

Of course you know that
\[\log(ax)=\log a+\log (x)\]
and due to log a is a constant, it gets out of the derivaitve of the main function

Answer 12

for ex:
\[\frac{ d }{ dx }(2x^2)=2\frac{ d }{ dx }(x^2)\]

Answer 13

I would apply the product rule theoretically but Symbolab told me to leave as is.
(My thought process)\[\frac{d}{dx}\left((\log{2})\times(x^{2}-3x+6)\right)\]and apply the Product Rule from there

Answer 14

Yet what I was told was to leave off the \(\log{2}\) as \(\log_{10}{2}\)

Answer 15

OOOOOOOOOOOOOOOHHHHHHHHHHHHHHH
You wrote it wrong,MAN

This is not fair.

Why you did not write it correctly?

Answer 16

I did?

Answer 17

The problem on my paper is written as\[\log{2}(x^{2}-3x+6)\]

Answer 18

Here is the entire Symbolab result: https://goo.gl/Y3Xlzx

Answer 19

You mean that
\[\frac{ d }{ dx }[(\log2)(x^2-3x+6)]=\frac{ d }{ dx }[constant*polynimial]\]
\[=constant\frac{ d }{ dx }[polynomial]\]

Answer 20

@kittiwitti1

Answer 21

Question: the tangent passes through (2,2) or tangent AT (2,2) ??

Answer 22

@3mar eh?

@Loser66 "Find an equation to the tangent line to the curve at the given point"

Answer 23

and like @3mar question. I would like to know which one is correct one
\(y =log (2x^2-6x+12)\) (1)

or \(y= (x^2-3x+6) log2\) (2)
Which one??

Answer 24

(2)

Answer 25

so, it is simple enough. :)

Answer 26

I don't know. That's what the paper says.

Answer 27

I told you!

Answer 28

?

Answer 29

but!!

Answer 30

I applied the Product Rule as multiplying two forms together.

Answer 31

(2,2) is not on the curve, how can you find the tangent at (2,2)?

Answer 32

You don't need to apply the Product Rule.
Just derive the polynomial

Answer 33

Passes through (2,2) , we have a way to find it out.

Answer 34

Derive polynomial? @3mar

Answer 35

Yes.
It is simple now

Answer 36

I think you are right @Loser66 the tangent line PASSES through (2,2)

Answer 37

OH, because it is not \(\log_{b}{x}\) but rather\(\log_{b}{a}\), yes?

Answer 38

Thus not a function of \(x\)

Answer 39

\[\frac{ dy }{ dx }=(\log2)*\frac{ dy }{ dx }(x^2-3x+6)=\log2(2x-3)\]
apply (2,2)
\[\frac{ dy }{ dx }|_{x=2}=\log2(2*2-3)=(\log2)*(1)=\log2=0.301\]

Answer 40

Yes you are right

Answer 41

Why does the constant stay outside? Isn't derivative of a constant 0...
D: I'm confused

Answer 42

NO MAN
Because it is multiplied by the polynomial
\[\frac{ d }{ dx }(constant*x^2)=constant*\frac{ d }{ dx }(x^2)\]

Answer 43

log 2 is constant not a function multiplied by a function.

Answer 44

But it is within the y-function (equation?) o.o

Answer 45

Y-function may contain constants in the polynomial.

ex:
\[y=x^2+12x+36\]
16 is a constant

Answer 46

16?\[y'=\left(x^{2}+12x+36\right)'\rightarrow(2x+12(1)+0)=2x+12\]

Answer 47

I get the feeling I am applying the wrong logic to this

Answer 48

You got it!

Answer 49

But that would make log 2 = 0 by derivative constant rule ... ?

Answer 50

... I DO apply the Product Rule, don't I :\

Answer 51

"But that would make log 2 = 0 by derivative constant rule ... ?"
This is in case of dy/dx(log2+x^2-34+6) not dy/dx((log 2)(x^2-34+6))
got it?

Answer 52

\[\left(\log{2}(x^{2}-3x+6)\right)'=(\log{2})'•(x^{2}-3x+6)+\log{2}\left(x^{2}-3x+6\right)'\]\[=0(x^{2}-3x+6)+\log{2}(2x-3(1)+0)=0+\log2(2x-3)=\log2\times(2x-3)\]

Answer 53

Does that work? D:

Answer 54

Yes. This is the applying the rule literally.
Great job!
we can deduce from this the rule:
\[\frac{ d }{ dx }(5x^3)=5\frac{ d }{ dx }(x^3)=5(3x^2)\]

Answer 55

Oh, okay. I was utterly confused for the longest time lol
Thank you! @3mar

Answer 56

But now?

Answer 57

Thank you for the medal!

Answer 58

I'm good now. ☺

Answer 59

Thanks for Allah
I am happy to hear that!

Answer 60

Thank you for the medal!