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Expert-verified Found in: Page 132 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # If $$V$$ is the volume of a cube with edge length $$x$$ and the cube expands as time passes, find $$\frac{{dV}}{{dt}}$$ in terms of $$\frac{{dx}}{{dt}}$$.

The value of $$\frac{{dV}}{{dt}}$$ in terms of $$\frac{{dx}}{{dt}}$$ is $$\left( {{\rm{3}}{x^{\rm{2}}}} \right)\frac{{{\rm{d}}x}}{{{\rm{dt}}}}$$.

See the step by step solution

## Step 1: Definition of the Chain rule

If $$y = f\left( u \right)$$ and $$u = g\left( x \right)$$ are differentiable functions, then the Chain rule is given by $$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}$$

## Step 2: Find $$\frac{{dV}}{{dt}}$$ in terms of $$\frac{{dx}}{{dt}}$$.

It is given that a cube of a volume V with edge length $$x$$.

Then, the volume of the cube is given by, $$V\left( x \right) = {x^3}$$.

Differentiate both sides of $$V\left( x \right) = {x^3}$$ with respect to $$t$$ as follows:

$$\frac{{dV}}{{dt}} = \frac{d}{{dt}}\left( {{x^3}} \right)$$

Apply the chain rule of differentiation as follows:

$$\begin{array}{c}\frac{{dV}}{{dt}} = \frac{d}{{dx}}\left( {{x^3}} \right)\frac{{dx}}{{dt}}\\\frac{{dV}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}\end{array}$$

Therefore, $$\frac{{dV}}{{dt}}$$ in terms of $$\frac{{dx}}{{dt}}$$ is equal to $$3{x^2}\frac{{dx}}{{dt}}$$. ### Want to see more solutions like these? 