Question

Use any method to evaluate the derivative of the following functions. $$h(x)=\left(5 x^{7}+5 x\right)\left(6 x^{3}+3 x^{2}+3\right)$$

Short answer

Question: Find the derivative of the function $$h(x)=\left(5 x^{7}+5 x\right)\left(6 x^{3}+3 x^{2}+3\right)$$ Answer: The derivative of the function h(x) is $$\frac{dh(x)}{dx} = 300x^9 + 135x^8 + 105x^6 + 120x^3 + 45x^2 + 15$$

Step-by-step solution

Identify the functions u(x) and v(x)

Let the function \(u(x) = 5x^7 + 5x\) and \(v(x) = 6x^3 + 3x^2 + 3\).

Find the derivative of u(x) and v(x)

Now we need to find the derivatives of u(x) and v(x). Using the power rule for derivatives, we have: $$\frac{du}{dx} = 35x^6 + 5$$ $$\frac{dv}{dx} = 18x^2+6x$$

Apply the product rule

Now that we have the derivatives of both functions, we can apply the product rule: $$\frac{dh(x)}{dx} = \left(5x^7 + 5x\right)\left(18x^2+6x\right) + \left(6x^3 + 3x^2 + 3\right)\left(35x^6+5\right)$$

Simplify the expression (Optional)

If you wish to simplify the expression, you can expand the product and group like terms: $$\frac{dh(x)}{dx} = 5x^7(18x^2) + 5x^7(6x) + 5x(18x^2) + 5x(6x) + 6x^3(35x^6) + 6x^3(5) + 3x^2(35x^6) + 3x^2(5) + 3(35x^6) + 3(5)$$ After simplifying, we get: $$\frac{dh(x)}{dx} = 90x^9 + 30x^8 + 90x^3 + 30x^2 + 210x^9 + 30x^3 + 105x^8 + 15x^2 + 105x^6 + 15$$ Now, combining like terms, we get the final answer: $$\frac{dh(x)}{dx} = 300x^9 + 135x^8 + 105x^6 + 120x^3 + 45x^2 + 15$$

Key concepts

Product Rule

Understanding the product rule is fundamental in calculus when dealing with the differentiation of products of two or more functions. When you have a function expressed as the product of two other functions—say, \( u(x) \) and \( v(x) \)—the derivative of the product is not simply the product of their derivatives. The product rule states that to find this derivative, \( \frac{d}{dx}[u(x)v(x)] \), you use the formula \( \frac{d}{dx}[u(x)v(x)] = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx} \).

In our exercise, \( u(x) \) was \( 5x^7 + 5x \) and \( v(x) \) was \( 6x^3 + 3x^2 + 3 \). By applying the product rule, we expressed \( \frac{dh(x)}{dx} \) as a sum of two terms: one where \( u(x) \) is multiplied by the derivative of \( v(x) \) and another where \( v(x) \) is multiplied by the derivative of \( u(x) \)—exactly as the rule suggests. This rule is crucial because it allows the differentiation of complex functions that are the products of simpler functions.

Power Rule for Derivatives

When you're faced with the task of differentiating polynomials or functions with variables raised to a power, the power rule for derivatives is an indispensable tool. It states that if \( f(x) = x^n \), where \( n \) is a constant, then the derivative of \( f \) with respect to \( x \) is \( f'(x) = nx^{n-1} \). This makes differentiating polynomials a straightforward process.

In our original problem, we differentiated \( u(x) \) and \( v(x) \) using the power rule. For \( u(x) = 5x^7 + 5x \), its derivative is \( 35x^6 + 5 \) because you bring down the exponent as a coefficient and reduce the exponent by one. The same process was applied to \( v(x) \) to find \( 18x^2 + 6x \) as its derivative. This rule is not only helpful but also a time-saver when working through calculus problems involving powers.

Simplifying Expressions

After applying differentiation rules, we often need to simplify the resulting expressions to make them more manageable and to identify any patterns or further relationships. Simplification can involve expanding products, combining like terms, and reducing fractions. It requires attentiveness to algebraic manipulation and is vital for presenting your final answer in the simplest form possible.

In the solution we discussed, simplifying the derivative after applying the product rule involved a number of steps. First, we expanded the product of each term in the derivatives of \( u(x) \) and \( v(x) \) to get a longer expression involving multiple terms with various powers of \( x \). Then, we combined terms with the same power of \( x \) which is essentially grouping like terms together. This process is critical as it transforms the expression into a reduced, more understandable form and, in this case, showed how many terms canceled or combined to form the neatly packaged final answer.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's a tool that allows us to study change and motion and is divided mainly into two subfields: differential calculus and integral calculus. Differential calculus concerns itself with the concept of the derivative, which we've explored in our exercise. It enables us to find the rate at which quantities change, which is essential in various fields like physics, engineering, economics, and beyond.

Our exercise focused on differential calculus, particularly on taking derivatives of functions. As seen in the problem-solving steps, understanding the principles of calculus—like the product rule and power rule—can simplify complex problems into solvable parts. Calculus concepts such as these have numerous applications and are foundational for advancing in any scientific or mathematical field.