Question

Last question of the night!
find the limit as h approaches 0:
[cos(x+h)^7 -cosx^7]/h

Answer 1

This may seem simple, but I just can't figure out where to start. Brain, don't quit on me now!

Answer 2

It's multiple choice:
a. -7x^6sinx^7
b. -7cos^6xsinx
c. -sin7x^13
d. infinity
e. 0

Answer 3

@phi are you still on? could you please help me once more?

Answer 4

I know I can't use direct substitution because that produces indeterminate form. yes.
but I don't think they want me to factor or simplify, either.
am I supposed to do something with differentiation here?

Answer 5

can you use L'Hospital's Rule ?

Answer 6

I've heard of that! but strangely, it doesn't come up as a lesson until a few units later, at least. I wasn't sure if that is how they wanted me to solve it then.
I have never worked with this rule. could you please help with that?

Answer 7

it's like, differentiate numerator and then differentiate denominator and then take the limit, right?

Answer 8

yes. you use it when you have an indeterminate form such as 0/0 (the case here)
or infinity/infinity

Answer 9

ok so how should I go about it when there are variables x and h?
(oh dear, is that a question I should know the answer to?) >_<

Answer 10

can we just identify this as the definition of the derivative of cos^7(x), or do we have to do this the long painful way?

Answer 11

in this case, we treat x as a constant

Answer 12

btw, just to make sure this is
\[ \cos\left( (x+h)^7\right) \]?

Answer 13

well, I can't say for sure, but i'm leaning towards long and painful.
yes phi, it is

Answer 14

so then we are differentiating with respect to h?
is that what you mean by treating x as a constant?

Answer 15

maybe it is better to keep track of x and get factors of dx/dh
they will subtract out...

Answer 16

if you look at the form, it's the definition of the derivative of cos(x^7), so you could just observe that and take the derivative wrt x. I really wouldn't know how to approach the problem otherwise, except to start hacking away with various trig identities or l'hospital.

If anyone has a more rigorous proof I'd sure like to see it though :D

Answer 17

I think we were going to use l'hospital.
i'm still a bit confused though.

Answer 18

turingtest's idea makes sense. We could use l'hospital, but if you have not studied it yet, maybe they want you to say:

that is the definition of the derivative of cos(x^7). and find the derivative of cos(x^7)

Answer 19

ok that does make sense
but for the second part, are you saying that the whole equation is just an expanded definition of the derivative of cos(x^7)?
so we would just take the derivative of that?
(would we use chain rule?)
u= cosx
y= u^7
?

Answer 20

cos(u)
u = x^7

Answer 21

so derivative is:
-sin(u)?
-sinx^7?

Answer 22

or would you take derivative of u before substituting back in?
7x^6
-7sinx^6?

Answer 23

@phi
are you there?
which of those two would it be?

Answer 24

d cos(u)= -sin(u) du
u is x^7
du = d x^7 = 7 x^6

we get
- sin u du = -sin(x^7) 7 x^6 or
-7 x^6 sin(x^7)

Answer 25

oh, okay! I missed a part.
so then our answer is a!
and no need for the l'hospital rule!

Answer 26

no need for l'hospital

btw, when I say
d cos(u) that is short of d/dx cos (u)
= -sin(u) du/dx

when we get to d/dx x^7 we get
7 x^6 dx/dx or just 7 x^6

Answer 27

right, ok!
I believe that's it, then! I'll go over this once more to make sure I get it.
thanks so much for your help today!
i'm closing the question.