Related Rates / Implicit Differentiation

**millerst** · April 5th 2009, 07:16 AM

I'm so lost with this question:

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

Can someone help me with this question please?

**e^(i*pi)** · April 5th 2009, 07:19 AM

Originally Posted by millerst

I'm so lost with this question:

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

Can someone help me with this question please?

You can use the chain rule:

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

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

**skeeter** · April 5th 2009, 08:06 AM

Originally Posted by millerst

I'm so lost with this question:

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

Can someone help me with this question please?

$\frac{d}{dt}(y = x^2 + 4x)$

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

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

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

**millerst** · April 5th 2009, 08:10 AM

Sketter, what happened to the dy/dx?

**skeeter** · April 5th 2009, 08:24 AM

Originally Posted by millerst

Sketter, what happened to the dy/dx?

there is no $\frac{dy}{dx}$ ... the derivative was taken with respect to $t$ ... that's what the $\frac{d}{dt}$ in front of the equation means.

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