chain rules

**iwonder** · October 15th 2008, 03:29 PM

z=x^2y+xy^2
x=5+t^5
y=-2+t^2
how to find dz/dt?

**skeeter** · October 15th 2008, 03:43 PM

$\frac{dz}{dt} = x^2\frac{dy}{dt} + 2xy\frac{dx}{dt} + 2xy\frac{dy}{dt} + y^2\frac{dx}{dt} = x^2\frac{dy}{dt} + 2xy\left(\frac{dx}{dt} + \frac{dy}{dt}\right) + y^2\frac{dx}{dt}$

$x = 5+t^5$

$\frac{dx}{dt} = 5t^4$

$y = t^2 - 2$

$\frac{dy}{dt} = 2t$

if you want $\frac{dz}{dt}$ in terms of $t$ , perform the substitutions.

**iwonder** · October 15th 2008, 04:01 PM

10t+20t^3-80t^4+56t^6+4t^8-20t^9+10t^11is it the right answer?

**skeeter** · October 15th 2008, 04:15 PM

I don't know ... I left the grunt work algebra to you.

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