Derivative of f(x) = sin(2x) by limit definition

[MarkSA]

Hello,

I have this problem. I'm wondering if someone can provide a few tips about how to solve this to steer me in the right direction.

f(x) = sin(2x)
Use the definition of a derivative to solve for f'(x).

I put it in form:
lim h->0, [sin(2x+2h) - sin(2x)]/h
and proceeded to apply all of the double angle and sum formulas. I'm left with a wonderfully gigantic mess that I can't seem to do anything with. (I was able to use the lim of sinh/h = 1 rule a few times but nothing more.)

I can't seem to get it into a form where I can factor an h out of the numerator to cancel out the h in the denominator in order to set the rest of the h's to 0 and solve. Any suggestions?

Edit: I can isolate a lim h->0 of [cos^2h-1]/h.. if I could set that to 0 it would help matters immensely. But the rule only works for [cosh-1]/h, correct?

 

[daon]

Re: Derivative by definition

Hello,

I have this problem. I'm wondering if someone can provide a few tips about how to solve this to steer me in the right direction.

f(x) = sin(2x)
Use the definition of a derivative to solve for f'(x).

I put it in form:
lim h->0, [sin(2x+2h) - sin(2x)]/h
and proceeded to apply all of the double angle and sum formulas. I'm left with a wonderfully gigantic mess that I can't seem to do anything with. (I was able to use the lim of sinh/h = 1 rule a few times but nothing more.)

I can't seem to get it into a form where I can factor an h out of the numerator to cancel out the h in the denominator in order to set the rest of the h's to 0 and solve. Any suggestions?

There's always a trick. Trying to figure out which things to substitute in for and which ones to not.

\(\displaystyle \L

\frac{ \sin\( 2x+2h\)-\sin\(2x\)}{h} \,\, = \,\
\frac{\sin\(2x\)\cos\(2h\)+\cos\(2x\)\sin\(2h\)-\sin\(2x\)}{h} \,\,\)

As h goes to zero, 2h goes to zero and cos(0)=1. So,...

\(\displaystyle \frac{\sin\(2x\)\cos\(0\)+\cos\(2x\)\sin\(2h\)-\sin\(2x\)}{h} \,\, =
\frac{\sin\(2x\)+\cos\(2x\)\sin\(2h\)-\sin\(2x\)}{h} \,\, = \,\,
\frac{\cos\(2x\)\sin\(2h\)}{h} \,\,\)

Then one last manipulation.. As you said, use the \(\displaystyle \,\, \lim_{x \rightarrow 0} \,\, \frac{\sin\(x\)}{x} =1\) rule...

\(\displaystyle \frac{\cos\(2x\)\sin\(2h\)}{h} \,\, = \,\, \cos\(2x\)\(\frac{\sin\(2h\)}{h}\) = \,\, 2\cos\(2x\)\(\frac{\sin\(2h\)}{2h}\) = 2cos(2x)\)

 

[MarkSA]

Hi daon,

Thanks for the reply. Your method is a surprise to me. We were basically taught that it was a 'no-no' to set h's to 0 at any point if it would make the denominator undefined. I say h's plural because we were taught that setting the h's to 0 was roughly the last step in the process and that all were zeroed at once. (i've never encountered a problem before where it was necessary to set one individual h to 0)

This has been the primary roadblock for me in solving this problem.

Is there perhaps another way to do this by the method I was taught, factoring out an h somehow to cancel the one in the denominator (even though the way you did it clearly results in the correct answer)? I think i'm a little confused over the selective setting of a single h to zero. (For instance, sin(2h) couldn't be set to sin(0) since that would result in a limit of 0)

 

[daon]

Re: Derivative by definition

Hi daon,

Thanks for the reply. Your method is a surprise to me. We were basically taught that it was a 'no-no' to set h's to 0 at any point if it would make the denominator undefined. I say h's plural because we were taught that setting the h's to 0 was roughly the last step in the process and that all were zeroed at once. (i've never encountered a problem before where it was necessary to set one individual h to 0)

This has been the primary roadblock for me in solving this problem.

Is there perhaps another way to do this by the method I was taught, factoring out an h somehow to cancel the one in the denominator (even though the way you did it clearly results in the correct answer)? I think i'm a little confused over the selective setting of a single h to zero. (For instance, sin(2h) couldn't be set to sin(0) since that would result in a limit of 0)

Actually if you set sin(2h) to zero you'd get zero, but your end result would be 0/0. I find no error in putting in zero for h if it doesn't cause any problems. In any case, heres a way you may like better.



\(\displaystyle \L
\frac{\sin\(2x\)\cos\(2h\)+\cos\(2x\)\sin\(2h\)-\sin\(2x\)}{h} \,\, =

\frac{\sin\(2x\)\cos\(2h\)-\sin\(2x\)}{h}+\frac{\cos\(2x\)\sin\(2h\)}{h} \,\,\)

Playing with the fractions a bit we get

\(\displaystyle \L
\sin\(2x\)\(\frac{\cos\(2h\)-1}{h}\)+\cos\(2x\)\frac{\sin\(2h\)}{h} \,\, =
2\sin\(2x\)\(\frac{\cos\(2h\)-1}{2h}\)+2\cos\(2x\)\(\frac{\sin\(2h\)}{2h}\)\)

Then using your cos, sin limits, you have...

\(\displaystyle \L
2\sin\(2x\)\(\frac{\cos\(2h\)-1}{2h}\)+2\cos\(2x\)\(\frac{\sin\(2h\)}{2h}\) =
\sin\(2x\)\(0\)+2\cos\(2x\)\(\1\) = 2\cos\(2x\)\)

 

[MarkSA]

Yes, that way makes much more sense to me based on how I was taught to do this type of problem! (Though i'll have to keep your trick in mind too, as that would have made this a whole lot easier.)

Thanks.

 

[daon]

Yes, that way makes much more sense to me based on how I was taught to do this type of problem! (Though i'll have to keep your trick in mind too, as that would have made this a whole lot easier.)

Thanks.

No problem, although I would listen to your teacher in doing these problems. (S)He's the one grading your work!