**Proof:**

To prove that the function

d/x sinh(u) = cosh(u) * du/dx

we can use the chain rule of differentiation.

**Chain Rule:**

The chain rule states that if we have a function y = f(g(x)), then the derivative of y with respect to x can be found by multiplying the derivative of f with respect to g, and the derivative of g with respect to x.

Mathematically, if y = f(g(x)), then dy/dx = df/dg * dg/dx.

**Applying the Chain Rule:**

Let's apply the chain rule to prove the given function.

Consider the function y = sinhx, where u = g(x) is the inner function.

We can rewrite the function as y = sinhu, where u = g(x).

Now, let's differentiate both sides of the equation with respect to x.

d/dx (y) = d/dx (sinhu)

By the chain rule, we have:

dy/dx = dy/du * du/dx

**Differentiating y = sinhu:**

To find dy/du, we differentiate y = sinhu with respect to u.

dy/du = coshu (derivative of sinh(u) with respect to u is cosh(u))

**Differentiating u = g(x):**

To find du/dx, we differentiate u = g(x) with respect to x.

du/dx = d/dx (g(x))

**Combining the Results:**

Now, we can substitute the values of dy/du and du/dx into the chain rule equation:

dy/dx = dy/du * du/dx

dy/dx = coshu * du/dx

Substituting y = sinhu, we have:

d/dx (sinhu) = coshu * du/dx

Therefore, we have proved that:

d/x sinhu = coshu * du/dx

This demonstrates that the given function is true based on the chain rule of differentiation.

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