# Thread: Integral Differentiation

1. ## Integral Differentiation

Can anyone please help with the following equation? (file attached).

The 3xy2 is confusing me.

Thanks

2. ## implicit differentiation

Originally Posted by Beginner
Can anyone please help with the following equation? (file attached).

The 3xy^2 is confusing me.

Thanks
product rule. let the first function be $\displaystyle 3x$ , the second $\displaystyle y^2$ ...

$\displaystyle 3x \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 3$

3. Originally Posted by Beginner
Can anyone please help with the following equation? (file attached).

The 3xy2 is confusing me.

Thanks
Hi! You can split $\displaystyle 3xy^2$ into $\displaystyle 3x$ times $\displaystyle y^2$ and use the product rule:

$\displaystyle \frac{d}{dx} (3x)(y^2) = 3(y^2) + (3x) \left(2y \frac{dy}{dx} \right)$

4. Originally Posted by Beginner
Can anyone please help with the following equation? (file attached).

The 3xy2 is confusing me.

Thanks
The equation is
$\displaystyle x^3 + 3xy^2 + 4y = 40$
After differentiation you get
$\displaystyle 3x^2 + 3x(2y)(dy/dx) + 3y^2 + 4dy/dx = 0$
Now simplify to find dy/dx

5. Originally Posted by skeeter
product rule. let the first function be $\displaystyle 3x$ , the second $\displaystyle y^2$ ...

$\displaystyle 3x \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 3$
Thanks for your help its much appreciated.

6. Originally Posted by mathemagister
Hi! You can split $\displaystyle 3xy^2$ into $\displaystyle 3x$ times $\displaystyle y^2$ and use the product rule:

$\displaystyle \frac{d}{dx} (3x)(y^2) = 3(y^2) + (3x) \left(2y \frac{dy}{dx} \right)$
Thanks Mathemagister!

7. Originally Posted by sa-ri-ga-ma
The equation is
$\displaystyle x^3 + 3xy^2 + 4y = 40$
After differentiation you get
$\displaystyle 3x^2 + 3x(2y)(dy/dx) + 3y^2 + 4dy/dx = 0$
Now simplify to find dy/dx
Thanks for the help sa-ri-ga-ma