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This is an integral in my lecture and they said that the chain rule is used here to produce the second equation.

I understand the chain rule with single variables however I am baffled here and how the chain rule applies to multivariable equations since I don't think we can 'cancel out' anything due to partial derivatives being different to derivatives.

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So the normal chain rule is $$ \frac{d}{dx} f(g(x)) = f'(g(x))g'(x)$$

In words: the rate of change of $f(g(x))$ in terms of $x$ is equal to the rate of $f$ in terms of $g(x)$ times the rate of change of $g$ in terms of $x$.

Now we extend this idea to multiple variables: If $$ \phi(g(x), y, z) $$ then $$ \frac{\partial}{\partial x} \phi(g(x), y, z) = \frac{\partial}{\partial g(x)} \phi(g(x), y, z) \cdot g'(x)$$ In words: the rate of change of $\phi$ in terms of $x$ is equal to the rate of change of $\phi$ in terms of $g(x)$ (assuming $y$ and $z$ remain unchanged) times the rate of change of $g$ in terms of $x$.

Now in your example we want to differentiate $\phi(x(s), y(s), z(s))$ in terms of $s$. The same principle applies, only $x$, $y$ and $z$ all depend on $s$ so

$$ \frac{d}{ds} \phi(x(s), y(s), z(s)) = \frac{\partial\phi}{\partial x}\cdot x'(s) + \frac{\partial\phi}{\partial y}\cdot y'(s) + \frac{\partial\phi}{\partial z}\cdot z'(s)$$

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The chain rule for multivariate maps states that the Jacobian of a map that is the two composition of two maps is the (matrix) product of the Jacobians of two maps.

This is what is applied in the special case you mention in your question.

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I don't think we can 'cancel out' anything

Well 'cancel out' is not exactly what is going on. Leibniz notation is not actually a ratio; its an analogy.

However, this analogy does make the chain rule easier to remember.   So in the same manner, you may present the multivariate chain rule as a dot product of vectors.

$$\begin{align}\dfrac{\mathrm d \varphi}{\mathrm d s} &= \dfrac{\partial \varphi}{\partial\langle x,y,z\rangle}\cdot\dfrac{\mathrm d\langle x,y,z\rangle}{\mathrm d s}\\[2ex]&=\left\langle\dfrac{\partial\varphi}{\partial x},\dfrac{\partial\varphi}{\partial y},\dfrac{\partial\varphi}{\partial z}\right\rangle\cdot\left\langle\dfrac{\mathrm d x}{\mathrm d s}, \dfrac{\mathrm d y}{\mathrm d s},\dfrac{\mathrm d z}{\mathrm d s}\right\rangle\\[2ex]&=\dfrac{\partial\varphi}{\partial x}\dfrac{\mathrm d x}{\mathrm d s}+\dfrac{\partial\varphi}{\partial y}\dfrac{\mathrm d y}{\mathrm d s}+\dfrac{\partial\varphi}{\partial z}\dfrac{\mathrm d z}{\mathrm d s}\end{align}$$

due to partial derivatives being different to derivatives.

The partial differentiation of a monovariate function equals the differentiation, with respect to the argument.$$\dfrac{\partial f(s)}{\partial s}=\dfrac{\mathrm d f(s)}{\mathrm d s}$$

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