# Thread: solving problem using definition of derivative

1. ## solving problem using definition of derivative

$f(x) = x^3 + 2x$

Use the definition of the derivative as a limit to determine the slope of the
tangent line to f(x) at x = 1

I know the rule is limit as h goes to 0 $f(x+h) - f(x)\over {h}$

But how do I rearrange this one so the answer isn't 0?

2. this is called finding the limit

$\frac{(x+h)^3 +2(x+h) -x^3 -2x}{h}$

= $\frac{x^3 +3x^2h +3xh^2 + h^3 +2x +2h - x^3 -2x}{h}$

= $\frac{3x^2h +3xh^2 +h^3 +2h}{h}$

= $\frac{h(3x^2 +3xh +h^2 +2)}{h}$

= $3x^2 +3xh +h^2 +2$

Now plug in 0 for x and you get

$3x^2 +2$

3. Originally Posted by 11rdc11
this is called finding the limit

...
= $3x^2 +3xh +h^2 +2$

Now plug in 0 for h and you get <=== tiny typo

$3x^2 +2$

...
$f(x) = x^3 + 2x$

Use the definition of the derivative as a limit to determine the slope of the
tangent line to f(x) at x = 1

I know the rule is limit as h goes to 0 $f(x+h) - f(x)\over {h}$

But how do I rearrange this one so the answer isn't 0?
1. Using the formula you know that

$f'(x)=\lim_{h \to 0}\left(\dfrac{((x+h)^3+2(x+h))-(x^3+2x)}{h}\right)$

2. You are asked to calculate the slope of the graph at x = 1, that means:

$f'(1)=\lim_{h \to 0}\left(\dfrac{((1+h)^3+2(1+h))-(1^3+2\cdot 1)}{h}\right)$

3. Expand the brackets in the numerator:

$f'(1)=\lim_{h \to 0}\left(\dfrac{(1+3h+3h^2+h^3+2+2h)-(3)}{h}\right)$

4. Collect like terms and factor the numerator. Cancel the factor h. You'll get:

$f'(1)=\lim_{h \to 0}(h^2+3h+5) = 5$