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

#### MarkFL

Staff member
Here is the question:

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:

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
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: 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.

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#### Prove It

##### Well-known member
MHB Math Helper
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. Also, the interpretation I used is a somewhat more interesting problem. And a tangent line to a linear function seemed to be a bit strange. 