91Ó°ÊÓ

The power rule is a fundamental tool in calculus for finding derivatives. It allows you to differentiate expressions of the form \(x^n\). If you have a function \(f(x)=x^n\), the derivative \(f'(x)\) is found using the power rule, which states:

• \(\frac{d}{dx}(x^n) = n \cdot x^{n-1}\)

This means you multiply the power \(n\) by the coefficient in front of \(x\), and then subtract one from the exponent.
In our exercise, we start with \((5x)^{-3}\). Using the power rule, we get:

• \(\frac{d}{dx}((5x)^{-3}) = -3 \cdot (5x)^{-4}\)

So, the exponent -3 is brought down as a multiplier, and the new exponent is -4.

The constant multiple rule is another basic concept in calculus differentiation. It simplifies the process of differentiation when a constant is multiplied with a function. According to this rule, if you have \(c \cdot f(x)\), the derivative is:

• \(\frac{d}{dx}(c \cdot f(x)) = c \cdot \frac{d}{dx}f(x)\)

Essentially, the constant \(c\) stays as it is, and only the function \(f(x)\) is differentiated.
In our scenario, we apply this rule right from the start. Our function is rewritten as \(6 \cdot (5x)^{-3}\). When differentiating, the 6 remains untouched initially, and we differentiate \((5x)^{-3}\), which we've already done using the power rule.
So, the result becomes:

• \(6 \cdot (-3) \cdot (5x)^{-4}\)

By simplifying, the constant multiplication combines to give \(-18 \cdot (5x)^{-4}\).

The chain rule is a powerful derivative tool used when differentiating composite functions, which are functions within functions. It allows us to break down complex expressions into simpler parts. Given a composite function \(f(g(x))\), the chain rule tells us:

• \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

You differentiate the outer function and then multiply it by the derivative of the inner function.
In this exercise, the inner function is \(5x\) and the outer function is \((5x)^{-4}\) after applying the power rule. By applying the chain rule, the derivative becomes:

• \(\frac{d}{dx}((5x)^{-4}) = -4 \cdot (5x)^{-5} \cdot 5\)

The \(-4\) comes from the outer function's derivative, \((5x)^{-5}\) remains from the outer function, and multiplying by 5 accounts for the derivative of the inner function \(5x\).
Thus, it gives us the product \(-20 \cdot (5x)^{-5}\), simplifying the process throughout the differentiation.