Derivatives of a Quadratic Function with Step-by-Step Instructions

[mathaTon]

Hi
I need help with the Derivites please.


Find the value of dy/dx for the given value of x.

y= (3-2x-x^2) (x^2+x-2), x=-2

I have tried that I can't seem to understand the real steps to get the answer.
I know I have to take the deriviate of this actual function...
once i do that..I am not sure what to do next? please help!

 

[Office_Shredder]

Once you have the derivative, just plug in x.

Or are you having trouble taking the derivative?

 

[HallsofIvy]

y is a product of two functions, 3- 2x- x² and x²+ x- 2 so use the product rule: (fg)'= f 'g+ fg'. Or just go ahead and multiply it to get a single polynomial. Do you know how to find the derivative of a polynomial?

 

[mathaTon]

no i don't know how to take the derivative of a polynomial?

is that? let me try though.
first i expand this two terms:
i ended with some cubes nd quartics
then i took the derivative.(when u multiply the exponent with coffiencent and the subtract the exponent with 1)

thats wehre i get stuck. i need help here..

the middle step is difficult..
last part is easy when we plug x value..

 

[3trQN]

$$y = (-x^2-2x+3)(x^2+x-2)$$

let:
$$u = f(x) = (-x^2-2x+3)$$
$$v = g(x) = (x^2+x-2)$$
$$y = uv$$

$$y + dy = (u + du)(v + dv) = uv + udv + vdu + dudv$$

subtract y = uv
$$dy = udv + vdu$$

this equates to:
$$\frac{dy}{dx} = (-x^2-2x+3)\frac{d}{dx}(x^2+x-2) + (x^2+x-2)\frac{d}{dx}(-x^2-2x+3)$$
$$\frac{dy}{dx}=(-x^2-2x+3)(-2x-2) + (x^2+x-2)(2x+1)$$

Thats my attempt :P

EDIT: Corrected

 

[VietDao29]

$$\frac{dy}{dx} = (-x^2-2x+3)\frac{d}{dx}(-x^2-2x+3) + (x^2+x-2)\frac{d}{dx}(x^2+x-2)$$

Uhmm, this line is wrong. What you have written in this line is udu + vdv, in stead of udv + vdu.

no i don't know how to take the derivative of a polynomial?

Uhmm, you know that:
$$\left( x ^ {\alpha} \right) ' = \alpha x ^ {\alpha - 1}$$, right?
And you should also know that the derivative of a sum, is the sum of derivatives, i.e:
(u + v + w + k) = u' + v' + w' + k'
And if k is a constant then:
[kf(x)]' = k f'(x)
A polynomial has the form:
$$n_m x ^ m + n_{m - 1} x ^ {m - 1} + ... + n_1 x + n_0$$
This is a polynomial of degree m.
The derivative of that polynomial is:
$$\left( n_m x ^ m + n_{m - 1} x ^ {m - 1} + ... + n_1 x + n_0 \right)' = n_m \left( x ^ m \right)' + n_{m - 1} \left( x ^ {m - 1} \right)' + ... + n_1 (x)' + (n_0)'$$
$$= mn_m \left( x ^ {m - 1} \right)' + (m - 1)n_{m - 1} \left( x ^ {m - 2} \right)' + ... + n_1$$.

first i expand this two terms:
i ended with some cubes nd quartics
then i took the derivative.(when u multiply the exponent with coffiencent and the subtract the exponent with 1)

Uhmm, not very sure what you mean though. Can you type it out?
--------------------
The problem ask you to find the derivative of a product. You can use the Product Rule:
(uv)' = u'v + uv'
By letting:
u = 3 - 2x - x²
and v = x² + x - 2.
We have:
[(3 - 2x - x²) (x² + x - 2.)]' = (3 - 2x - x²)' (x² + x - 2.) + (3 - 2x - x²) (x² + x - 2.)'.
Can you go from here? :)

 

[3trQN]

Sorry, that because of my poor latex skills more than my maths