Two variables, x and y, are related by the equation
$\displaystyle y=\frac{3}{4}(\frac{x}{12}-1)^6.$
Given that both x and y vary with time, find the value of y when the rate of change of y is 12 times the rate of change of x.
And, just in case a picture helps fill out the chain rule...
... where straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case time), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
Starting then with...
Spoiler:
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
$\displaystyle y = \frac{3}{4}\left(\frac{x}{12} - 1\right)^6$
$\displaystyle \frac{dy}{dt} = \frac{9}{2}\left(\frac{x}{12} - 1\right)^5 \cdot \frac{1}{12} \cdot \frac{dx}{dt}$
$\displaystyle 12 \cdot \frac{dx}{dt} = \frac{9}{2}\left(\frac{x}{12} - 1\right)^5 \cdot \frac{1}{12} \cdot \frac{dx}{dt}$
assuming $\displaystyle \frac{dx}{dt} \ne 0$ ...
$\displaystyle 12 = \frac{9}{2}\left(\frac{x}{12} - 1\right)^5 \cdot \frac{1}{12}$
$\displaystyle 12 = \frac{3}{8}\left(\frac{x}{12} - 1\right)^5$
$\displaystyle 32 = \left(\frac{x}{12} - 1\right)^5$
$\displaystyle 2^5 = \left(\frac{x}{12} - 1\right)^5$
$\displaystyle x = 36$
calculate the value of y from the original equation.