Compute Derivative Using First Principles - MathLoveR's Question

[MarkFL]

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.

 

[MarkFL]

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:

https://www.physicsforums.com/attachments/826._xfImport

To MaThLoVeR and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

 

[Prove It]

Are you sure that \(\displaystyle \displaystyle f(x) = \frac{4}{3x}\) and not \(\displaystyle \displaystyle \frac{4}{3}x\)?

 

[MarkFL]

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)

 

[CellCoree]

Hello MathLoveR,

Thank you for your question regarding computing derivatives using first principles. I am happy to assist you with this problem.

a) To compute f'(3), we will use the definition of derivative which is the limit of the difference quotient as h approaches 0. This can be written as:

f'(3) = lim(h->0) (f(3+h) - f(3)) / h

Plugging in the given function, we get:

f'(3) = lim(h->0) ((4/3(3+h)) - (4/3(3))) / h

= lim(h->0) ((4/3(3+h) - 4) / h

= lim(h->0) (4/3h) / h

= lim(h->0) 4/3

= 4/3

Therefore, f'(3) = 4/3.

b) Now, to find the equation of the line tangent to the curve y = f(x) at the point where x = 2, we will use the result from part (a). The equation of a line can be written as y = mx + b, where m is the slope and b is the y-intercept.

Since we know the slope at x = 3 is 4/3, we can plug that into the equation as:

y = (4/3)x + b

To find the y-intercept, we can plug in the point (2, f(2)) into the equation:

f(2) = (4/3)(2) + b

= 8/3 + b

Now, since we know that f(2) = 4/3(2) = 8/3, we can substitute that in and solve for b:

8/3 = 8/3 + b

b = 0

Therefore, the equation of the line tangent to the curve y = f(x) at the point where x = 2 is y = (4/3)x.

I hope this helps you with your question. Keep up the good work in your calculus studies!

Best regards,