Derivative and tangent-line questions

[jumpman21]

1.) Let f'(x) = x/2x+3. Find f'(x) by taking the limit, as h goes to zero, of [ f(a+h) - f(a) ] / h

My steps are:

(a+h)/2(a+h) +3 - a/2a+3

(a+h)/2a +2h + 3 - a/2a+3

I need to find a common denominato. Should I multiply by 2h...? What steps follow?

2.) Algebraically find the equation of the tangent line to f(x) = x^(1/3) at (8, 2)

Completely stuck on this one. Point slope formula maybe? Thanks for help in advance.

 

[galactus]

Re: Derivitive and tangent line question.

Two questions.

1.) let f'(x) = x/2x+3, find f'(x) by f(a+h) - f(a)
---------------
h

My steps are:
(a+h)/2(a+h) +3 - a/2a+3

(a+h)/2a +2h + 3 - a/2a+3

I need to find a common denominator, should i multiply by 2h or what? What steps follow?

Please use grouping symbols. I assume you mean \(\displaystyle \L\\\frac{x}{2x+3}\).

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{x+h}{2(x+h)+3}-\frac{x}{2x+3}}{h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{3}{(2x+3)(2x+2h+3)}\)

Now, finish?.

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2.) Algebraically find the equation of the tangent line to f(x)=x^(1/3) at (8,2)

Completely stuck on this one. Point slope formula maybe? Thanks for help in advance.

Find the derivative of f(x) and plug in x to find the slope at that point. Use y=mx+b. You have m, x, and y. Plug them in and solve for b. You're done.

 

[jumpman21]

How did you get from the first step to the second one? How did you just invert 3 and put it over (2x+3)(2x+2h+3)

 

[stapel]

How did you get from the first step to the second one?

Try working it out yourself: What does "x + h - x" simplify to? How would [h/(stuff)]/h reduce? And so forth.

Eliz.