Product rule, calculus and simplifying

[Henrik]

Task is to differentiate the function with respect to x.

5x^2(2-x^3)^2

This is my attempt

u=5x^2 v=(2-x^3)^2

du/dx=10x dv/dx= 2(2-x^3)(-3x^2)= -6x^2(2-x^3)

dy/dx=u*dv/dx + v*du/dx = 5x^2 * -6x^2(2-x^3) + 10x(2-x^3)^2

= -30x^4(2-x^3) + 10x(2-x^3)^2

10x[ 3x^3 + (2-x^3)]

The answer is supposed to be

20x(2-x^3)(1-2x^3)

I dont understand what I have done incorrectly.

Solved the one above and got a new problem. The method is still the product rule.

x^2 * sqrt (3 - 4x)

My attempt

u= x^2 V= sqrt (3 - 4x)

du/dx = 2x dv/dx = -4 / 2 sqrt( 3 - 4x)

dy/dx = du/dx + dv/dx = -4 x^2 / 2 sqrt (3 - 4x) + 2x sqrt (3 - 4x)

= - 2x^2/ sqrt (3 - 4x) + 2x sqrt (3 - 4x)

The answer is 2x (3 - 5x) / sqrt (3 - 4x) Where have I done wrong?

 

[Bob Brown MSEE]

This is a simple poly. No product rule necessary.
(d/dx)(5xxxxxxxx-20xxxxx+20xx) = (40xxxxxxx-100xxxx+40x)

 

[MarkFL]

I am with you up to here: \(\displaystyle \dfrac{dy}{dx}=-30x^4(2-x^3)+10x(2-x^3)^2\)

You have simply factored incorrectly, after doing all the differentiation correctly.

Now, observe that both terms have \(\displaystyle 10x(2-x^3)\) as a factor:

\(\displaystyle \dfrac{dy}{dx}=10x(2-x^3)(-3x^3+(2-x^3))=10x(2-x^3)(2-4x^3)=20x(2-x^3)(1-2x^3)\)

 

[Henrik]

I am with you up to here: \(\displaystyle \dfrac{dy}{dx}=-30x^4(2-x^3)+10x(2-x^3)^2\)

You have simply factored incorrectly, after doing all the differentiation correctly.

Now, observe that both terms have \(\displaystyle 10x(2-x^3)\) as a factor:

\(\displaystyle \dfrac{dy}{dx}=10x(2-x^3)(-3x^3+(2-x^3))=10x(2-x^3)(2-4x^3)=20x(2-x^3)(1-2x^3)\)

Got it, thanks

 

[JeffM]

Task is to differentiate the function with respect to x.

5x^2(2-x^3)^2

This is my attempt

u=5x^2 v=(2-x^3)^2

du/dx=10x dv/dx= 2(2-x^3)(-3x^2)= -6x^2(2-x^3)

dy/dx=u*dv/dx + v*du/dx = 5x^2 * -6x^2(2-x^3) + 10x(2-x^3)^2 This is wrong

\(\displaystyle u = 5x^2 \implies u' = 10x\) You got this right

\(\displaystyle v = (2 - x^3)^2 \implies v' = 2(2 - x^3)(- 3x^2)\) You got this right

\(\displaystyle y = uv \implies y' = u * v' + u' * v\) And you got this right

\(\displaystyle So\ y' = 5x^2 * 2(2 - x^3)(-3x^2) + 10x(2 - x^3)^2\) When you did this you ignored the \(\displaystyle - 3x^2\) Careless error.

= -30x^4(2-x^3) + 10x(2-x^3)^2

10x[ 3x^3 + (2-x^3)]

The answer is supposed to be

\(\displaystyle y' = 5x^2 * 2(2 - x^3)(-3x^2) + 10x(2 - x^3)^2 = 10x(2 - x^3)\{x(-3x^2) + (2 - x^3)\} = 20x(2 - x^3)(1 - 2x^3).\) It is

20x(2-x^3)(1-2x^3)

I dont understand what I have done incorrectly.

Solved the one above and got a new problem. The method is still the product rule.

x^2 * sqrt (3 - 4x)

My attempt

u= x^2 V= sqrt (3 - 4x)

du/dx = 2x Good dv/dx = -4 / (2 sqrt( 3 - 4x)) Good except for the missing grouping symbols.

dy/dx = du/dx + dv/dx NO!!!!!!!!!!!!!! dy/dx = v(du/dx) + u(dv/dx)

= -4 x^2 / (2 sqrt (3 - 4x))+ 2x sqrt (3 - 4x) But you did it correctly here except for the grouping symbols

= - 2x^2/ sqrt (3 - 4x) + 2x sqrt (3 - 4x)

\(\displaystyle = \dfrac{-2x^2 + 2x\sqrt{3 - 4x}\sqrt{3 - 4x}}{\sqrt{3 - 4x}} = \dfrac{2x(3 - 4x - x)}{\sqrt{3 - 4x}} = \dfrac{2x(3 - 5x)}{\sqrt{3 - 4x}}\)

The answer is 2x (3 - 5x) / sqrt (3 - 4x) Where have I done wrong? You forgot to simplify

You OK now?