Question

Derivatives Find and simplify the derivative of the following functions. $$y=(2 \sqrt{x}-1)(4 x+1)^{-1}$$

Short answer

Question: Find and simplify the derivative of the function y = (2√x - 1)(4x + 1)^{-1}. Answer: The derivative of the function y is given by y' = (8 - 32√x) / [x^(1/2)(4x + 1)^2].

Step-by-step solution

Find the derivatives u'(x) and v'(x)

To find the derivative of u(x) = 2√x - 1, we have: $$u'(x)=\frac{d}{dx}(2\sqrt{x}-1)$$ $$u'(x)=\frac{d}{dx}(2x^{\frac{1}{2}}-1)$$ Now find the derivative of each term: $$u'(x)=2\frac{1}{2}x^{\frac{1}{2}-1}-0$$ $$u'(x)=x^{-\frac{1}{2}}$$ To find the derivative of v(x) = (4x + 1)^{-1}, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function: $$v'(x)=\frac{d}{dx}[(4x+1)^{-1}]$$ First, find the derivative of the outer function: $$\frac{d}{dx}[u^{-1}] = -u^{-2}$$ Next, find the derivative of the inner function: $$4$$ Finally, apply the chain rule: $$v'(x) = -u^{-2} * 4$$ $$v'(x) = -\frac{4}{(4x+1)^2}$$

Apply the product rule

Now that we have u'(x) and v'(x), we can apply the product rule to find the derivative of the given function: $$y' = u' * v + u * v'$$ Substitute the derivatives we found in Step 1: $$y' = x^{-\frac{1}{2}}(4x+1)^{-1} + (2\sqrt{x} - 1)\left(-\frac{4}{(4x + 1)^2}\right)$$

Simplify the derivative

Lastly, we will simplify the expression for the derivative: $$y' = \frac{1}{x^{\frac{1}{2}}(4x+1)} - \frac{4(2\sqrt{x} - 1)}{(4x+1)^2}$$ $$y' = \frac{(4x+1) - 8\sqrt{x}(4x+1) + 4}{x^{\frac{1}{2}}(4x+1)^2}$$ $$y' = \frac{4 - 32\sqrt{x} + 4}{x^{\frac{1}{2}}(4x+1)^2}$$ $$y' = \frac{8 - 32\sqrt{x}}{x^{\frac{1}{2}}(4x+1)^2}$$ Therefore, the derivative of the given function is: $$y' = \frac{8 - 32\sqrt{x}}{x^{\frac{1}{2}}(4x+1)^2}$$

Key concepts

Product Rule

One of the fundamental concepts in calculus is the product rule, which is essential for finding the derivative of functions that are products of two or more functions. In layman's terms, when you have a function that is made up of two distinct parts multiplied together, you can't just differentiate each part separately and multiply the results. Instead, you must apply the product rule.

The generic formula for the product rule is: \[y' = u'v + uv'\] Where, \(u\) and \(v\) are functions of \(x\), and \(u'\) and \(v'\) represent their derivatives. This rule essentially says that the derivative of the product of two functions is the derivative of the first times the second plus the first times the derivative of the second.

Using the product rule correctly is critical for ensuring that complex functions are differentiated accurately. This rule showcases how intertwined the parts of a product function are with respect to their rates of change - a concept which can reflect various real-world scenarios, such as the relationship between velocity and acceleration.

Chain Rule

Another indispensable tool in calculus is the chain rule, used for taking the derivative of a composite function – that is, a function that's nested inside another function. The chain rule is somewhat analogous to opening a set of Russian dolls, wherein each doll represents a function tucked inside another.

The chain rule can be stated as: \[ \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \] This powerful rule describes that to find the derivative of the composite function \(f(g(x))\), we first differentiate the outer function \(f\) with respect to its inner function \(g(x)\), then multiply that by the derivative of the inner function \(g\) with respect to \(x\).

For example, if a function is raised to a power, like \((4x + 1)^{-1}\), we first find the derivative as if the inner part, \(4x + 1\), stood alone, then multiply by the derivative of the inner part. The chain rule is indispensable when dealing with exponential functions, trigonometric functions, and any function where one quantity depends on another.

Simplifying Derivatives

Finally, simplifying derivatives is a crucial step that shouldn't be overlooked. Simplification can turn a complex, daunting expression into a more manageable and understandable form. Once you've applied rules like the product and chain rules, you'll often end up with a lengthy and complicated derivative. Rewriting and reducing this derivative makes further analysis and interpretation easier.

To simplify derivatives effectively:

• Combine like terms wherever possible.
• Factor expressions to reveal common factors that can be canceled out.
• Use algebraic identities to condense expressions.
• Simplify complex fractions by multiplying by the reciprocal, if necessary, or finding a common denominator.

In our exercise, after applying the product rule, the simplification process involved combining terms over a common denominator, and then reducing the expression to its simplest form. This clarity is especially useful when calculating critical points of a function or solving applied problems. By following a clear and methodical approach, we can reveal the simplest form of a derivative - an insight that often exposes the heart of calculus problems.