Simplified differentiation?

Question

Using differentiation rules how do i differentiate the following to its most simplest form?

Y=(6x^2 + 2x)^3

and

y=x^2 + 3x^3 / loge 2x

log e (e being an subscript e)

Answer 1

Y=(6x^2 + 2x)^3

using the chain rule...
y' = 3(6x^2 + 2x)^2 (12x + 2)

In the second one... by Loge, do you mean "natural log"?

natural log (ln) = log base e, so I'm going to use "ln" since it's easier to look at.

y = (x^2 + 3x^3) / (ln 2x)

The quotient rule states:
derivative of f(x)/g(x)
= (g*f ' - f*g ') / g^2

f(x) = x^2 + 3x^3
so
f ' (x) = 2x + 9x^2

g(x) = ln 2x
so
g ' (x) = 2(1/2x) = 1/x

So
y ' = [(ln 2x)(2x + 9x^2) - (x^2 + 3x^3)(1/x)] / (ln 2x)^2

you can simplify it a little bit...
y' = (2x + 9x^2) / (ln 2x) - (x + 3x^2) / (ln 2x)^2

Answer 2

I could help you with the first one, but the second one includes logs, which i dont learn in connection to differentiation untill next year-

for the first one you would first simplify the equation. Since you are multiplying the powers, you need to multiply a power of 3 to the existing x powers-

ie. y=6x^2x3 + 2x^1x3 ~ y=6x^6 + 2x^3

Then you would differentiate this new equation. To differentiate, you times the power of the x by the integar (number) before the x, and then reduce the power by 1.

Therefore-

dy/dx= 36x^5 + 6x^2

you could then simplfy this if necessary to get~ 6x^5 + x^2

Hope this helped you, sorry i couldn't help you with the second one!

Answer 3

Y = 2*6x + 3*2x = 12x+6x = 18x is the dirivative of the first one .

Use the division rule for taking derivatives to solve the second one in the calculas book .
then put in the values for your variables in the derivatives
to evaluate it .

Answer 4

Question 1
dy/dx = 3.(6x² + 2x)² X (12x + 2)
dy/dx = 3.(12x + 2).4x².(3x + 1)²
dy/dx = 6.(6x + 1).4x².(3x + 1)²
dy/dx = 24.x².(6x + 1).(3x + 1)²
OR
let u = 6x² + 2x
du/dx = 12x + 2 = 2.(6x + 1)
y = u³
dy/du = 3u²
dy/dx = dy/du X du/dx
dy/dx = 3.(6x² + 2x)² X 2.(6x + 1)
dy/dx = 3.4x².(3x + 1)² X 2.(6x + 1)
dy/dx = 24x².(6x + 1).(3x + 1)²

Question 2
Consider u = 3x³ / log 2x
Quotient rule:-
du/dx = (log2x.9x² - 6x³.log2x) / (log2x)²
dy/dx is then:-
[2x.(log2x)² + (log2x.9x² - 6x³.log 2x)] / (log2x)²
[ 2x.(log2x) + 3x².(3 - 2x) ] / (log2x)