Welcome to our community

Be a part of something great, join today!

MaThLoVeR's question at Yahoo! Answers regarding computing a derivative using first principles

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

Calculus Question .... Please HELP?

For the function f(x) = 4/3x

a) Using the definition of derivative (limits) to compute f'(3)
b) Use the result of part (a) to find an equation of the line tangent to the curve y = f(x) at the point for which x = 2
Here is a link to the question:

Calculus Question .... Please HELP? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello MaThLoVeR,

We are given:

\(\displaystyle f(x)=\frac{4}{3x}\)

a) To compute the derivative of $f$ using first principles, we use:

\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)

Using the given function definition, we have:

\(\displaystyle f'(x)=\lim_{h\to0}=\frac{\frac{4}{3(x+h)}-\frac{4}{3x}}{h}\)

In the numerator of the expression, let's combine the two terms by getting a common denominator:

\(\displaystyle f'(x)=\lim_{h\to0}=\frac{\frac{4(3x)}{3(x+h)(3x)}-\frac{4(3(x+h)}{3x(3(x+h))}}{h}\)

\(\displaystyle f'(x)=\lim_{h\to0}=\frac{4(3x)-4(3(x+h)}{3h(x+h)(3x)}\)

Distribute in the numerator:

\(\displaystyle f'(x)=\lim_{h\to0}=\frac{12x-12x-12h}{3h(x+h)(3x)}\)

Combine like terms:

\(\displaystyle f'(x)=\lim_{h\to0}=\frac{-12h}{3h(x+h)(3x)}\)

Divide out common factors:

\(\displaystyle f'(x)=\lim_{h\to0}=\frac{-4}{(x+h)(3x)}=-\frac{4}{3x^2}\)

Hence:

\(\displaystyle f'(3)=-\frac{4}{3(3)^2}=-\frac{4}{27}\)

b) Now, to find the tangent line where $x=2$, we need a point on the curve:

\(\displaystyle (2,f(2))=\left(2,\frac{2}{3} \right)\)

and we need the slope:

\(\displaystyle m=f'(2)=-\frac{4}{3(2)^2}=-\frac{1}{3}\)

Using the point-slope formula, we find the equation of the tangent line is:

\(\displaystyle y-\frac{2}{3}=-\frac{1}{3}(x-2)\)

Arranging in slope-intercept form, we have:

\(\displaystyle y=-\frac{1}{3}x+\frac{4}{3}\)

Here is a plot of the function and the tangent line:

mathlover.jpg

To MaThLoVeR and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Are you sure that \(\displaystyle \displaystyle f(x) = \frac{4}{3x}\) and not \(\displaystyle \displaystyle \frac{4}{3}x\)?
 
  • Thread starter
  • Admin
  • #4

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Are you sure that \(\displaystyle \displaystyle f(x) = \frac{4}{3x}\) and not \(\displaystyle \displaystyle \frac{4}{3}x\)?
No, in fact I am not sure since bracketing symbols were not used, so I made a guess as to what was intended. (Bandit)

Also, the interpretation I used is a somewhat more interesting problem. :D

And a tangent line to a linear function seemed to be a bit strange.(Rofl)