Derivatives: Tangent Line

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reardear

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Find an equation of the tangent line to the graph of the function

f(x)=x1/2(9-4x2) / x

where x=9. I plugged in 9 to get the points (9,-105) and then tried to solve for f'9 and got 350/27.

Then, to get the equation, I plugged in for y-y1=m(x-x1) and got y=(350/27)x-(665/3) in ax+b format. I'm not sure whether I'm correct or not, but these large numbers don't look too good to me. Can I please get some help with the steps to solving this question so I can try it out?


PS. How can I format equations here to show them properly?
 
reardear said:
Find an equation of the tangent line to the graph of the function

f(x)=x1/2(9-4x2) / x

where x=9. I plugged in 9 to get the points (9,-105) and then tried to solve for f'9 and got 350/27.

Then, to get the equation, I plugged in for y-y1=m(x-x1) and got y=(350/27)x-(665/3) in ax+b format. I'm not sure whether I'm correct or not, but these large numbers don't look too good to me. Can I please get some help with the steps to solving this question so I can try it out?


PS. How can I format equations here to show them properly?
Click to expand...

I am getting f'(9) = - 109/6 .................... so check your work.

There is no lawthat says the numbers in a problem cannot be messy. In real problem, those are always messy. That's why calculators and computers were invented.

Your approach to the problem is correct - do not worry about messy numbers.
 
Detailed work:

f(x)=x1/2(9-4x2) / x

f(x) = 9 * x-1/2 - 4 * x3/2

f'(x) = 9 (-1/2)* x-3/2 - 4 (3/2) * x 1/2

f'(9) = -18.1667 = -(18 + 1/6) = - 109/6
 
Subhotosh Khan said:
Detailed work:

f(x)=x1/2(9-4x2) / x

f(x) = 9 * x-1/2 - 4 * x3/2

f'(x) = 9 (-1/2)* x-3/2 - 4 (3/2) * x 1/2

f'(9) = -18.1667 = -(18 + 1/6) = - 109/6
Click to expand...

Yup. I had an error in my calc. I deleted my previous post so as to not confuse anyone.
 
Subhotosh Khan said:
Detailed work:

f(x)=x1/2(9-4x2) / x

f(x) = 9 * x-1/2 - 4 * x3/2

f'(x) = 9 (-1/2)* x-3/2 - 4 (3/2) * x 1/2

f'(9) = -18.1667 = -(18 + 1/6) = - 109/6
Click to expand...
What about using this formula?
f(x+h)-f(x) / h

That's what I learned and used, but your answer is still right. I'm confused now :(
 
reardear said:
What about using this formula?
f(x+h)-f(x) / h

That's what I learned and used, but your answer is still right. I'm confused now :(
Click to expand...

Using that formula would be UGLY! That formula is a way to arrive at the derivative using the method of limits. If your teacher wanted you to use that method, I feel for you brotha'.
 
reardear said:
What about using this formula?
f(x+h)-f(x) / h

That's what I learned and used, but your answer is still right. I'm confused now :(
Click to expand...
Not quite sure where you are in your studies.

\(\displaystyle f'(x) = \displaystyle \lim_{h \rightarrow 0}\dfrac{f(x + h) - f(x)}{h}\ is\ a\ definition.\)

But there are a number of theorems derived from that definition that make determining the derivative much easier than using the definition.

Example.

\(\displaystyle f(x) = m(x) + n(x) \implies \displaystyle f'(x) =\lim_{h \rightarrow 0}\dfrac{f(x + h) - f(x)}{h} =\)

\(\displaystyle \displaystyle \lim_{h \rightarrow 0}\dfrac{m(x + h) + n(x + h) - [m(x) + n(x)]}{h} = \lim_{h \rightarrow 0}\dfrac{[m(x + h) - m(x)] + [n(x + h) - n(x)]}{h} =\)

\(\displaystyle \displaystyle \lim_{h \rightarrow 0}\dfrac{m(x + h) - m(x)}{h} + \lim_{h \rightarrow 0}\dfrac{n(x + h) - n(x)}{h} = m'(x) + n'(x).\)

In short, \(\displaystyle f(x) = m(x) + n(x) \implies f'(x) = m'(x) + n'(x).\)

Probably you will start learning those theorems very quickly because as Sir Michael says, working from the definition is a pain. Of course you have to know how to use the definition when you do not know an applicable theorem.
 
Last edited:
Yep, it really is annoying to use, so thanks. I'm about to learn the chain rule in the next lecture, so I might have already gone over those theorems - I don't recognize them though. Then again... all I can do in class is focus on quickly copying everything down before the professor goes on to the next page. I'll go over my notes.

Thanks for your help guys!

Edit: Jeff, how do you type down equations like that?
 
reardear said:
Yep, it really is annoying to use, so thanks. I'm about to learn the chain rule in the next lecture, so I might have already gone over those theorems - I don't recognize them though. Then again... all I can do in class is focus on quickly copying everything down before the professor goes on to the next page. I'll go over my notes.

Thanks for your help guys!

Edit: Jeff, how do you type down equations like that?
Click to expand...
First, you're welcome.

Second, this site supports LaTex, which is a text editing language. I doubt that it is worth burdening your mind with learning LaTex to ask a few questions. You can get an idea what it looks like by hitting reply with quote to this post. However, it's much better when using this site to ask questions that are clear and to show your work right up to the point where you are stuck or unsure.

Third, the chain rule is an important theorem, particuarly because it allows you to combine the results of other theorems, but it usually comes after you have learned several other theorems, all easily derivable from the fundamental definition that you have been working with.

\(\displaystyle f(x) = a,\ where\ a\ is\ a\ constant \implies f'(x) = 0.\)

\(\displaystyle f(x) = m(x) + n(x) \implies f'(x) = m'(x) + n'(x).\) Addition Rule

\(\displaystyle f(x) = m(x) * n(x) \implies f'(x) = \{m'(x) * n(x)\} + \{n'(x) * m(x)\}.\) Multiplication Rule

\(\displaystyle f(x) = m(x) - n(x) \implies f'(x) = m'(x) - n'(x).\)

\(\displaystyle f(x) = \dfrac{m(x)}{n(x)} \implies f'(x) = \dfrac{\{m'(x) * n(x)\} - \{n'(x) * m(x)\}}{\{n(x)\}^2}.\) Quotient Rule

\(\displaystyle f(x) = ax^n,\ where\ n\ is\ an\ integer \implies f'(x) = anx^{n-1}.\) Power Rule
 
JeffM said:
First, you're welcome.

Second, this site supports LaTex, which is a text editing language. I doubt that it is worth burdening your mind with learning LaTex to ask a few questions. You can get an idea what it looks like by hitting reply with quote to this post. However, it's much better when using this site to ask questions that are clear and to show your work right up to the point where you are stuck or unsure.

Third, the chain rule is an important theorem, particuarly because it allows you to combine the results of other theorems, but it usually comes after you have learned several other theorems, all easily derivable from the fundamental definition that you have been working with.
Click to expand...
Thanks! I found this site to be helpful :D http://webdemo.visionobjects.com/equation.html?locale=default


mmm4444bot said:
You have not yet learned the Power Rule?
Click to expand...
Yeah I learned it the day after I posted this :p
 
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