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When differentiating functions in calculus, one crucial tool is the **Product Rule**. It helps when you have a function that is the product of two other functions. For the function \( h(x) = x^2 \arctan 5x \), we apply the Product Rule since \( h(x) \) is a product of \( u(x) = x^2 \) and \( v(x) = \arctan 5x \). The formula for the Product Rule is:

• \( (uv)' = u'v + uv' \)

To find \( h'(x) \), you first calculate the derivatives \( u'(x) = 2x \) and \( v'(x) \) – which is derived by using rules for trigonometric derivatives that we will explain in the next section. Then, substitute these into the Product Rule formula to get the derivative of the product.

Understanding trigonometric derivatives is essential for solving calculus problems involving trigonometric functions. In our expression \( \arctan 5x \), we need to find the derivative. The rule to find the derivative of \( \arctan u \) is:

• \( \frac{du}{dx} \frac{1}{1+u^2} \)

Here, \( u = 5x \), so \( u' = 5 \). Plug in \( u \) and \( u' \) to get: \( v'(x) = \frac{5}{1+(5x)^2} \). This step is critical as it simplifies handling more complex trigonometric functions in derivatives.

After you apply the Product Rule and find the derivatives as \( h'(x) = 2x (\arctan 5x) + x^2 \frac{5}{1+(5x)^2} \), you should consider if any simplification is possible. Simplifying derivatives helps to clean up expressions and sometimes reveals further insights or makes them easier to work with. In this particular case, the solution doesn’t allow for simple algebraic simplifications. It’s valuable to revisit complex terms to ensure there are no overlooked factors that could be reduced. Otherwise, as witnessed here, sometimes the most simplified form is the original answer you attained using calculus rules.

Calculus is the branch of mathematics where derivatives play a central role. It combines all the concepts we discussed and enables solving real-world problems involving changing quantities. By understanding derivatives, you can analyze functions to find rates of change, slopes of curves, and optimizations. In the exercise example, applying calculus principles like the Product Rule and trigonometric derivatives allows us to evaluate the differentiation of more complex functions, thus illustrating the essence and power of calculus. Developing a strong grasp of these concepts is key to succeeding in advanced mathematics and its applications in fields such as physics, engineering, and economics.