Thread: I really need help on derivatives

1. I really need help on derivatives

Part I
a.) find the slope of the tangent lineto the graph of the function f at point $(x_1,f(x_1))$. b.) Find the points on the graph where the tangent line is horizontal and use these points to sketch the graph.
$f(x) = 2x^3 - 3x^2$

Part II
Find the indicated derivative:
$D_x(\frac{1}{x^2} - x)$

Hey can you use the
$\lim_{h\to_ 0} \frac{f(x+h) - f(x)}{h}$ to solve the formula? coz i can really get it there

Part III
Find $\frac{dy}{dx}:$

$y= \frac{4}{2x-5}$

2. Unless you are asked to use the limit definition of the derivative, why not just use the rules/properties?

Part I
a.) find the slope of the tangent lineto the graph of the function f at point $(x_1,f(x_1))$. b.) Find the points on the graph where the tangent line is horizontal and use these points to sketch the graph.
$f(x) = 2x^3 - 3x^2$
Is this correct?
$a.) f(x_1) = 6x_1^2 - 6x_1$

Is my graph correct?
(1,-1) & (0,0) are the horizontal tangent lines

Is this correct?
$a.) f(x_1) = 6x_1^2 - 6x_1$

Is my graph correct?
(1,-1) & (0,0) are the horizontal tangent lines
Looks good to me.

-Dan

Part II
Find the indicated derivative:
$D_x(\frac{1}{x^2} - x)$
$\frac{d}{dx} \left ( \frac{1}{x^2} - x \right ) = -\frac{2}{x^3} - 1$

-Dan

Find $\frac{dy}{dx}:$
$y= \frac{4}{2x-5}$
$\frac{dy}{dx} = \frac{0 \cdot (2x-5) - 4(2)}{(2x-5)^2} = -\frac{8}{(2x-5)^2}$
2) $y = 4(2x-5)^{-1} = 4(-1)(2x-5)^{-2} \cdot 2 = -8(2x-5)^{-2} = -\frac{8}{(2x-5)^2}$