How Do You Apply the Chain Rule to Differentiate h = 3x^2y^3 with Respect to t?

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h = 3x^2.y^3
find dh/dt, if x=1, and y=2
Also, dx/dt = 0.2, dy/dt = 0.1

Any ideas where i should start in order to get this out?
Thanx
 
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\frac{dh}{dt} = \sum_i \frac{dF}{dx_i}\frac{dx_i}{dt}, you have the expression for h and the points at which to calculate the final answer.
 
I don't understand that expression, can you simplify your explanation please
 
If x is a function of t, f(x)= x^2, what does the chain rule say
df/dt is?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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