Comments for Steve the Math Tutor http://www.themathtutor.co.nz Mon, 13 Mar 2017 19:10:57 +0000 hourly 1 https://wordpress.org/?v=4.1.32 Comment on Q&A by Steve http://www.themathtutor.co.nz/qna/#comment-121 Mon, 13 Mar 2017 19:10:57 +0000 http://www.themathtutor.co.nz/?page_id=29#comment-121 You use both. Using long division gives the result you already have: x+1-1/(x^3-x). I haven’t double-checked this result for calculation errors but it’s certainly of the correct form.

Then the fraction 1/(x^3-x) can be simplified using partial fractions.

]]> Comment on Q&A by Anonymous http://www.themathtutor.co.nz/qna/#comment-119 Sat, 11 Mar 2017 08:12:49 +0000 http://www.themathtutor.co.nz/?page_id=29#comment-119 Hi, I’m confused about a problem: Use long division and partial fractions to simplify the rational function (x^4+x^3-x^2-x+1)/(x^3-x).

I’m confused on whether I should use partial fractions or long division, I get different answers for each method. I have also tried doing partial fractions and then long division and I get the answer: x+1-(1/(x^3-x)). Is this correct?

Thanks

]]> Comment on Q&A by Steve http://www.themathtutor.co.nz/qna/#comment-17 Wed, 22 Apr 2015 02:48:56 +0000 http://www.themathtutor.co.nz/?page_id=29#comment-17 Use Chain Rule. A similar term not requiring implicit differentiation would be: If \boldsymbol{y=\sin(3x^2)}, then

\boldsymbol{\frac{\text{d}y}{\text{d}x}=\cos(3x^2)\times (3x^2)'=\cos(3x^2)\times 6x=6x\cos(3x^2)}

The only difference with your problem, is that you have to employ implicit differentiation to find the derivative of \boldsymbol{xy}. But if you’re ok with differentiating \boldsymbol{y\cos(x)} that shouldn’t be a problem.

Hope that helps. I’ll write a larger post in the blog on implicit differentiation when I get the opportunity.

]]> Comment on Q&A by that guy with long hair who studies engineering http://www.themathtutor.co.nz/qna/#comment-16 Wed, 22 Apr 2015 00:32:06 +0000 http://www.themathtutor.co.nz/?page_id=29#comment-16 im getting stuck on the sin(xy) of this implicit differentiation problem.

ycos(x) = 1 +sin(xy)

could anyone help me?

]]> Comment on Q&A by Steve http://www.themathtutor.co.nz/qna/#comment-13 Mon, 16 Mar 2015 08:41:00 +0000 http://www.themathtutor.co.nz/?page_id=29#comment-13 Hi Totally Not Steve.

To solve the equation you need to get \boldsymbol{x} by itself on one side of the equation. This is my solution:

\boldsymbol{3x+4=19}
\boldsymbol{3x=15}
\boldsymbol{x=5}

Let me know if you’re not sure what I did at each step.

]]> Comment on Q&A by Totally Not Steve http://www.themathtutor.co.nz/qna/#comment-12 Mon, 16 Mar 2015 08:33:37 +0000 http://www.themathtutor.co.nz/?page_id=29#comment-12 Hi, I’m totally not Steve creating an example question on the Q&A page.

How do I solve 3x+4=19?

]]> Comment on Now with added Maths by Brian http://www.themathtutor.co.nz/2015/02/20/now-with-added-maths/#comment-9 Sat, 21 Feb 2015 19:38:16 +0000 http://www.themathtutor.co.nz/?p=48#comment-9 I’m viewing from the stock broeser on an android phone and I like it.

]]> Comment on Now with added Maths by Anonymous http://www.themathtutor.co.nz/2015/02/20/now-with-added-maths/#comment-7 Fri, 20 Feb 2015 21:48:19 +0000 http://www.themathtutor.co.nz/?p=48#comment-7 I’m completely anonymous and I like it too.

]]> Comment on Now with added Maths by Fharras Bigby http://www.themathtutor.co.nz/2015/02/20/now-with-added-maths/#comment-6 Fri, 20 Feb 2015 21:46:34 +0000 http://www.themathtutor.co.nz/?p=48#comment-6 I’m logged in and I like it!

]]> Comment on Now with added Maths by steve http://www.themathtutor.co.nz/2015/02/20/now-with-added-maths/#comment-2 Fri, 20 Feb 2015 00:32:09 +0000 http://www.themathtutor.co.nz/?p=48#comment-2 Sweet

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