# Derivative of $3xy$

I am confused im trying to get the derivative of $3xy$ out of this entire equation $$y= 4x^5 + 3xy + 5$$

Do I use the product rule here?

When I take the derivative, I try product rule.

$$(3)(y) \times (3x)(1)$$

So then I get $$3y \times 3x$$

Am I on the right track?

Maybe I am lost but I am assuming its product rule unless its another tactic I have to use here?

-
@soniccool Kindly use upper case "I" instead of lower case "i" to denote self. Using lower case "i" is incorrect. english.stackexchange.com/questions/172/… I edited few of your post to change lower case "i" to upper case "I". – user17762 Nov 22 '12 at 20:23

It is $3x \frac{dy}{dx}+3 y$ if you are using implicit differentiation.
I assume you want $\dfrac{dy}{dx}$. We have $$y = 4x^5 + 3xy + 5$$ Hence, $$\dfrac{dy}{dx} = \dfrac{d(4x^5 + 3xy + 5)}{dx} = \dfrac{d(4x^5)}{dx} + \dfrac{d(3xy)}{dx} + \dfrac{d(5)}{dx} = 20x^4 + \dfrac{d(3xy)}{dx}$$ Using product rule, we have that $$\dfrac{d(3xy)}{dx} = 3 \dfrac{d(xy)}{dx} = 3 \left(x \dfrac{dy}{dx} + \dfrac{dx}{dx} \cdot y \right) = 3 \left(y + x \dfrac{dy}{dx} \right)$$ Hence, we get that $$\dfrac{dy}{dx} = 20x^4 +3y + 3x \dfrac{dy}{dx} \implies \dfrac{dy}{dx} \left( 1-3x\right) = 20x^3 +3y \implies \dfrac{dy}{dx} = \dfrac{20x^4 +3y}{1-3x}$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}