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The derivative of a function is a fundamental concept in calculus that represents the rate at which a function changes. In simpler terms, it's like finding the slope of a curve at any given point. Mathematically, if you have a function \(f(x)\), its derivative is denoted as \(f'(x)\) or \(\frac{df}{dx}\). The derivative can be thought of as the "instantaneous rate of change" or the "slope of the tangent line" at a particular point on the graph of the function.

For the function \(g(x) = x^2 + f(x)\), the derivative is computed as:

\[g'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(f(x)) = 2x + f'(x)\]

Similarly, if we consider the function \(h(x) = 3f(x)\), its derivative is given by:

\[h'(x) = 3 \cdot f'(x)\]

This gives us an insight into how the function's output value changes as we change \(x\), giving us a powerful tool to analyze functions.

The chain rule is an essential tool in calculus to compute the derivative of composite functions. It allows us to find how a change in one variable affects another variable indirectly related through a function. To put it simply, if you have a function nested inside another function, the chain rule is applied.

Here, we used the chain rule while differentiating \(f(x)\) in both \(g(x)\) and \(h(x)\). The chain rule states:

• If \(z = f(g(x))\), then \(\frac{dz}{dx} = f'(g(x)) \cdot g'(x)\).

In this problem, during the derivative of \(g(x) = x^2 + f(x)\), we used:

• \(g'(x) = 2x + f'(x)\)

This is because \(f(x)\) is the inner function nested in \(g(x)\), and its derivative \(f'(x)\) was taken as part of the chain rule application. The rule simplifies complex derivatives and is crucial for solving calculus problems effectively.

The point-slope form is a mathematical formula used to find the equation of a line when you know its slope and a point on the line. This form is invaluable when writing the equation for tangent lines, which are lines that touch a curve at a single point without crossing it.

The formula for the point-slope form of a line is:

\[y - y_1 = m(x - x_1)\]

where \((x_1, y_1)\) is a known point on the line, and \(m\) is the slope.
In this example:

• The point for \(g(x)\) is \((3, 10)\) with a slope of 10, resulting in the tangent line equation:
  \[y - 10 = 10(x - 3)\]

After rearranging, it becomes:
\[y = 10x - 20\]

• For \(h(x)\), the point is \((3, 3)\) with a slope of 12:
• \[y - 3 = 12(x - 3)\]

This points to the importance of the point-slope form in finding linear equations quickly and efficiently.

Solving calculus problems often involves a series of ordered steps to break down complex information into manageable parts. Calculus problem solving requires a methodical approach:

• Identify the problem and what is being asked.
• Determine the functions involved and which calculus rules apply (e.g., chain rule, product rule).
• Compute the derivatives as necessary.
• Use the derivatives and given values to solve for specific outputs or equations like tangent lines.

In this problem's context, problem-solving involved finding tangent lines for the functions \(g(x)\) and \(h(x)\). The derivative of these functions was used to find their slopes at a particular point. Then:

• Using basic algebra (point-slope form), the equations of the tangent lines were deduced.

Such a structured approach helps in organizing thoughts and ensures accuracy in calculations, leading to the right solution.