Problem 100 Differentiate. $$F(x)=\left[6 ... [FREE SOLUTION] (2024)

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The product rule is essential when differentiating functions that are products of two simpler functions.
If you have a function that can be written as the product of two functions, say \f(x) \cdot g(x)\, the product rule states that the derivative is:
\( \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) \).
This rule helps by breaking down complicated expressions into simpler parts that are easier to handle.
In our exercise, we use the product rule to differentiate \[6x(3-x)^5\]. The steps include:

• Differentiating \f(x) = 6x\ to get \f'(x) = 6\.
• Differentiating \g(x) = (3-x)^5\ using the chain rule, resulting in \g'(x) = 5(3-x)^4 \cdot (-1) = -5(3-x)^4\.

Combining these, the derivative is:
\( 6(3-x)^5 + 6x \cdot (-5(3-x)^4) \)
After simplifying, we get \( 6(3-x)^5 - 30x(3-x)^4 \). See how the product rule makes things easier?

Now, let's put it all together using differentiation steps.
Start by identifying and differentiating your inner and outer functions.
For this problem, we found:

• \( g(u) = u^4 \) with \( g'(u) = 4u^3 \)
• \( u(x) = 6x(3-x)^5 + 2 \) with \( u'(x) = 6(3-x)^5 - 30x(3-x)^4 \)

We then use the chain rule that connects the derivatives of these functions:
\[F'(x) = g'(u) \cdot u'(x)\]
Substitute \( g'(u) \) and \( u'(x) \) back into this formula:
\[F'(x) = 4[6x(3-x)^5 + 2]^3 \cdot (6(3-x)^5 - 30x(3-x)^4)\]
This results in the final differentiated form of the function.
Breaking down complex differentiation into these steps ensures a clearer understanding.