What is the derivative of xy with respect to t, when x=8 and dx/dt=10?

homeylova223
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Hello everyone I have a two question


my first one is differentiate with respect to t

y=x^2-3x find dy/dt when x=3 and dx/dt=2

This is what I did
dy/dt=2x dx/dt (I am stuck right here)


my second one is

xy=4 find dy/dt when x=8 dx/dt=10
 
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homeylova223 said:
Hello everyone I have a two question


my first one is differentiate with respect to t

y=x^2-3x find dy/dt when x=3 and dx/dt=2

This is what I did
dy/dt=2x dx/dt (I am stuck right here)
What is the derivative of -3x with respect to t, in terms of dx/dt?


homeylova223 said:
my second one is

xy=4 find dy/dt when x=8 dx/dt=10

First, solve for y, using the first equation. Then get dy/dt in terms of x and dx/dt using the chain rule.
 
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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