Question

PLEASE HELP!!! MEDAL WILL BE REWARDED!
WHAT RULE DO I USE?
FIND Y':
y=3x^2+^3(sqrt x^4)

Answer 1

You use the rule

\[\large \frac{d}{dx}(x^n) = nx^{n-1}\]

and the idea that

\[\large \sqrt[n]{x^m} = x^{m/n}\]

Answer 2

So is that mean I would use product rule?

Answer 3

no because

\[\Large y = 3x^2 + \sqrt[3]{x^4}\]

is a sum (not a product)

Answer 4

oh so quotient?

Answer 5

you use the sum rule

\[\large \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]\]

Answer 6

basically: the derivative of the sum of two expressions/functions is the same as deriving the two functions individually first, then adding

Answer 7

oh okay, so I wold get dy dx =9x^2 x √ 3 +x 7/3

Answer 8

hold on, is the equation

\[\Large y = 3x^2 + \sqrt[3]{x^4}\]

OR is it

\[\Large y = 3x^2 * \sqrt[3]{x^4}\]

Answer 9

its +

Answer 10

ok, so you would do it like this

\[\Large y = 3x^2 + \sqrt[3]{x^4}\]

\[\Large y = 3x^2 + x^{4/3}\]

\[\Large \frac{d}{dx}[y] = \frac{d}{dx}[3x^2 + x^{4/3}]\]

\[\Large \frac{d}{dx}[y] = \frac{d}{dx}[3x^2] + \frac{d}{dx}[x^{4/3}]\]

\[\Large \frac{d}{dx}[y] = 3*2x^{2-1} + \frac{4}{3}x^{4/3-1}\]

\[\Large \frac{d}{dx}[y] = 3*2x^{1} + \frac{4}{3}x^{1/3}\]

\[\Large \frac{d}{dx}[y] = 6x + \frac{4}{3}x^{1/3}\]

\[\Large y^{\prime} = 6x + \frac{4}{3}x^{1/3}\]

Answer 11

ohhhh okay, im sorry I was using dx/dy instead of d/dx

Answer 12

you would use either dy/dx, f'(x), y' or d/dx[ ... ] (where you replace the dot dot dot with the expression)

dx/dy is going the wrong way

Answer 13

Oh okay, omg thank you very much!!!

Answer 14

you're welcome