My Math Forum 2 questions (limits/derivatives)

 Calculus Calculus Math Forum

 October 6th, 2012, 12:10 PM #1 Senior Member   Joined: Jun 2011 Posts: 154 Thanks: 0 2 questions (limits/derivatives) Hello everyone. I have a few problems I am stumped on. Studying for a test on Monday and need some guidance. Here are the problems: 1) Limit as x approaches 0 of (5x) / tan(7x). Can't plug in 0, as it will result in a 0 on the denominator. So I multiply by the reciprocal, which will be (5x) * [cos(x)/sin(x)]. And can not think of what to do next. Is it ok to plug the 0 into the denominator of the recriprocal, which would result in a 1 in the denominator, and leave us with 5x * cos(7x)? Or is that not the correct way to go about it. I don't know why but I always get stuck on these trig limit problems. 2) Given that f(x) = 3xh(x), h(2) = -1, and hprime(2) = -1 Find fprime(2) I know how to find derivatives I just don't know exactly how to go about this problem. Should I plug -1 in for x first? In fact while I'm at it I might as well ask for suggestions on the best way to go about solving this: find dy/dx of [(6x^2 - 1/x)]*(3x+1) I am familiar with the product rule, quotient rule, and with how to find derivatives. I just know that there are sometimes easier ways than others to find them. For the problem above, I end up with a mass of jumbled numbers, letters, and exponents. I am hoping someone can tell me how the simplest way to do it.
 October 6th, 2012, 12:21 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: 2 questions (limits/derivatives) 1.) $\lim_{x\to0}\frac{5x}{\tan(7x)}$ Let $u=7x\:\therefore\:u\to0$ $\lim_{u\to0}\frac{u}{\sin(u)}\cdot\frac{5}{7}\lim_ {u\to0}\cos(u)=\frac{5}{7}$ 2.) $f(x)=3xh(x)$ $f'(x)=3xh#39;(x)+3h(x)$ $f'(2)=3(2)h#39;(2)+3h(2)=6(-1)+3(-1)=-9$
October 6th, 2012, 12:47 PM   #3
Senior Member

Joined: Jun 2011

Posts: 154
Thanks: 0

Re: 2 questions (limits/derivatives)

Quote:
 Originally Posted by MarkFL 1.) $\lim_{x\to0}\frac{5x}{\tan(7x)}$ Let $u=7x\:\therefore\:u\to0$ $\lim_{u\to0}\frac{u}{\sin(u)}\cdot\frac{5}{7}\lim_ {u\to0}\cos(u)=\frac{5}{7}$
^^^ Thanks. This is a method I don't believe anyone has ever shown me. Is there a more commonly used method? Just so I can know in case I get stuck on another problem. If I multiply by the reciprocal (which would be 5x * cos7x / sin(7x) ), I am just wondering what would be the next step. Really I guess what I'm asking is how would you get the derivative of cos(7x) / sin(7x). I guess here I could just plug in 0, which would be cox(x) / 1, but is that correct?

Quote:
 2.) $f(x)=3xh(x)$ $f'(x)=3xh#39;(x)+3h(x)$ <<< Not sure what you did here. What happened to the x in the 2nd term? Why is the x gone but the h still there? $f'(2)=3(2)h#39;(2)+3h(2)=6(-1)+3(-1)=-9$
<<< So just take the derivative and then plug in 2 for x. Got it.

[/quote]

Not sure if you noticed but I did edit my post so if anyone can help me there is a 3rd problem in my original post I am also stuck on.

 October 6th, 2012, 01:12 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: 2 questions (limits/derivatives) If you know L'Hôpital's rule (most times it seems students haven't learned this yet, that's why I didn't use it), since you mention differentiation, then you could simply write: $\lim_{x\to0}\frac{5x}{\tan(7x)}=\lim_{x\to0}\frac{ 5}{7\sec^2(7x)}=\frac{5}{7}$ For your third problem, I would write: $y=$$6x^2-x^{\small{-1}}$$(3x+1)$ Now, using the product rule, we find: $\frac{dy}{dx}=$$6x^2-x^{\small{-1}}$$(3)+$$12x+x^{\small{-2}}$$(3x+1)=18x^2-3x^{\small{-1}}+36x^2+12x+3x^{\small{-1}}+x^{\small{-2}}=$ $54x^2+12x+x^{\small{-2}}=\frac{54x^4+12x^3+1}{x^2}$
October 6th, 2012, 01:30 PM   #5
Senior Member

Joined: Jun 2011

Posts: 154
Thanks: 0

Re: 2 questions (limits/derivatives)

Quote:
 Originally Posted by MarkFL If you know L'Hôpital's rule (most times it seems students haven't learned this yet, that's why I didn't use it), since you mention differentiation, then you could simply write: $\lim_{x\to0}\frac{5x}{\tan(7x)}=\lim_{x\to0}\frac{ 5}{7\sec^2(7x)}=\frac{5}{7}$
^^^ Can you tell me how and why the denominator, tan(7x), which has a derivative of sec^2(7x) (that I do know), but I don't know why it is multiplied by a 7 (the 7 in front of sec^2(7x) ). Is that another derivative process that I am not seeing? I've been known to get lost doing problems because I could never figure out when and where the chain rule was supposed to be applied.

Quote:
 For your third problem, I would write: $y=$$6x^2-x^{\small{-1}}$$(3x+1)$ Now, using the product rule, we find: $\frac{dy}{dx}=$$6x^2-x^{\small{-1}}$$(3)+$$12x+x^{\small{-2}}$$(3x+1)=18x^2-3x^{\small{-1}}+36x^2+12x+3x^{\small{-1}}+x^{\small{-2}}=$ $54x^2+12x+x^{\small{-2}}=\frac{54x^4+12x^3+1}{x^2}$

 October 6th, 2012, 01:36 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: 2 questions (limits/derivatives) Yes, the coefficient of 7 comes from the application of the chain rule as $\frac{d}{dx}(7x)=7$.
October 6th, 2012, 01:40 PM   #7
Senior Member

Joined: Jun 2011

Posts: 154
Thanks: 0

Re: 2 questions (limits/derivatives)

Quote:
 Originally Posted by MarkFL Yes, the coefficient of 7 comes from the application of the chain rule as $\frac{d}{dx}(7x)=7$.
Ok thanks. I feel more comfortable using that method for now.

Oh and I still am not quite sure about problem 2.

Not sure if you noticed I typed my message into the quote below:

Quote:
 2.) $f(x)=3xh(x)$ $f'(x)=3xh#39;(x)+3h(x)$ <<< Not sure what you did here. What happened to the x in the 2nd term? Why is the x gone but the h still there? $f'(2)=3(2)h#39;(2)+3h(2)=6(-1)+3(-1)=-9$
<<< So just take the derivative and then plug in 2 for x. Got it.

[/quote]

What process/rule was done to solve this problem? And also, as I noted above, in the 2nd part $f'(x)=3xh#39;(x)+3h(x)$ What happened to the x in the 2nd term? Why is the x gone but not the h?

 October 6th, 2012, 01:49 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: 2 questions (limits/derivatives) I used the product rule: $f'(x)=(3x)(h(x))'+(3x)'(h(x))=3xh#39; (x)+3h(x)$
October 6th, 2012, 01:56 PM   #9
Senior Member

Joined: Jun 2011

Posts: 154
Thanks: 0

Re: 2 questions (limits/derivatives)

Quote:
 Originally Posted by MarkFL I used the product rule: $f'(x)=(3x)(h(x))'+(3x)'(h(x))=3xh#39; (x)+3h(x)$
Gotcha. I guess I am having trouble telling where one function stops, and the other begins.

 October 6th, 2012, 04:21 PM #10 Senior Member   Joined: Jun 2011 Posts: 154 Thanks: 0 Re: 2 questions (limits/derivatives) ............. EDIT

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post unwisetome3 Calculus 3 October 20th, 2012 08:32 AM Ansh Agrawal Calculus 3 February 13th, 2012 07:25 AM alphabeta89 Real Analysis 12 January 29th, 2012 07:01 PM plm2e Real Analysis 1 September 23rd, 2008 11:51 AM unwisetome3 Algebra 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top