differentiate the function 𝑓 ( 𝑥 )

Let's differentiate the function $f(x) = x^3(7x^2 + 5)$ using the product rule.

Step-by-Step Solution:

The product rule states that for two functions $u(x)$ and $v(x)$, the derivative of their product is:

$\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

Here, we can set:

• $u(x) = x^3$
• $v(x) = 7x^2 + 5$

1. Differentiate $u(x) = x^3$:

$u'(x) = \frac{d}{dx}[x^3] = 3x^2$

2. Differentiate $v(x) = 7x^2 + 5$:

$v'(x) = \frac{d}{dx}[7x^2 + 5] = 14x$

3. Apply the product rule:

Now we apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute $u'(x) = 3x^2$, $v(x) = 7x^2 + 5$, $u(x) = x^3$, and $v'(x) = 14x$:

$f'(x) = (3x^2)(7x^2 + 5) + (x^3)(14x)$

4. Simplify each term:

$f'(x) = 3x^2(7x^2 + 5) + x^3(14x)$

$f'(x) = 3x^2(7x^2) + 3x^2(5) + x^3(14x)$

$f'(x) = 21x^4 + 15x^2 + 14x^4$

5. Combine like terms:

$f'(x) = (21x^4 + 14x^4) + 15x^2$

$f'(x) = 35x^4 + 15x^2$

Thus, the derivative of the function is: $f'(x) = 35x^4 + 15x^2$

Final Answer:

$f'(x) = 35x^4 + 15x^2$

Would you like further details or clarifications?

Related Questions:

1. How do you apply the product rule when there are more than two functions?
2. What happens if one of the functions is a constant when using the product rule?
3. Can the quotient rule be used instead of the product rule in this case?
4. How does the chain rule differ from the product rule?
5. What is the derivative of $g(x) = x^4(6x^3 + 2)$?

Tip:

Always simplify your functions first before differentiating if possible—it can make the process much easier!