The function given is $f(x) = x^6 e^x \cos(x)$, and we are tasked with finding its derivative $f'(x)$.

This expression involves a product of three functions:

• $x^6$,
• $e^x$,
• $\cos(x)$.

We will use the product rule for derivatives, which states that for functions $u(x)$, $v(x)$, and $w(x)$:

$(fg)' = f'g + fg'$

For three functions, this becomes: $(u \cdot v \cdot w)' = u' \cdot v \cdot w + u \cdot v' \cdot w + u \cdot v \cdot w'.$

Let:

• $u(x) = x^6$,
• $v(x) = e^x$,
• $w(x) = \cos(x)$.

Now, we compute the individual derivatives:

1. $u'(x) = \frac{d}{dx}(x^6) = 6x^5$,
2. $v'(x) = \frac{d}{dx}(e^x) = e^x$,
3. $w'(x) = \frac{d}{dx}(\cos(x)) = -\sin(x)$.

Now applying the product rule:

$f'(x) = u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x)$

Substituting the values: $f'(x) = 6x^5 \cdot e^x \cdot \cos(x) + x^6 \cdot e^x \cdot \cos(x) + x^6 \cdot e^x \cdot (-\sin(x))$

Simplify: $f'(x) = e^x \left( 6x^5 \cos(x) + x^6 \cos(x) - x^6 \sin(x) \right)$

Thus, the derivative is:

$f'(x) = e^x \left( (6x^5 + x^6) \cos(x) - x^6 \sin(x) \right).$

Would you like further details or clarifications?

Here are 5 related questions:

1. How would the derivative change if the function involved sine instead of cosine?
2. What happens if we add another function, say $\ln(x)$, to the original function?
3. Can you apply the product rule to more than three functions? How?
4. What is the second derivative of the function $f(x) = x^6 e^x \cos(x)$?
5. How does the behavior of the derivative change as $x$ approaches infinity?

Tip: Always look for ways to simplify expressions after applying the product rule; often terms can be factored for easier handling!