# Thread: Related Rates / Implicit Differentiation

1. ## Related Rates / Implicit Differentiation

I'm so lost with this question:

If $\displaystyle y = x^2 + 4x$ and dx/dt = 10 find dy/dt when x = 5.

Can someone help me with this question please?

2. Originally Posted by millerst
I'm so lost with this question:

If $\displaystyle y = x^2 + 4x$ and dx/dt = 10 find dy/dt when x = 5.

Can someone help me with this question please?
You can use the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

You know dt = 0.1 and dy/dx should be easy to find

3. Originally Posted by millerst
I'm so lost with this question:

If $\displaystyle y = x^2 + 4x$ and dx/dt = 10 find dy/dt when x = 5.

Can someone help me with this question please?
$\displaystyle \frac{d}{dt}(y = x^2 + 4x)$

$\displaystyle \frac{dy}{dt} = 2x\frac{dx}{dt} + 4\frac{dx}{dt}$

$\displaystyle \frac{dy}{dt} = \frac{dx}{dt}(2x+4)$

substitute in your given values for $\displaystyle x$ and $\displaystyle \frac{dx}{dt}$ to find the value for $\displaystyle \frac{dy}{dt}$

4. Sketter, what happened to the dy/dx?

5. Originally Posted by millerst
Sketter, what happened to the dy/dx?
there is no $\displaystyle \frac{dy}{dx}$ ... the derivative was taken with respect to $\displaystyle t$ ... that's what the $\displaystyle \frac{d}{dt}$ in front of the equation means.