Question

Group Activity In Exercises \(43-48,\) use the technique of logarithmic differentiation to find \(d y / d x\) . $$y=\sqrt[5]{\frac{(x-3)^{4}\left(x^{2}+1\right)}{(2 x+5)^{3}}}$$

Short answer

The derivative of the function \(y=\sqrt[5]{\frac{(x-3)^{4}\left(x^{2}+1\right)}{(2 x+5)^{3}}}\) with respect to x is \(\frac{dy}{dx} = \sqrt[5]{\frac{(x-3)^{4}\left(x^{2}+1\right)}{(2 x+5)^{3}}} \cdot \frac{1}{5} \left[\frac{4}{x-3} + \frac{2x}{x^2+1} - \frac{6}{2x+5}\right]\)

Step-by-step solution

Apply natural log

In this step, apply natural logarithm on both sides of the equation. This simplifies the given function making differentiation easier. So you get:\[\ln(y) = \ln\left(\sqrt[5]{\frac{(x-3)^{4}\left(x^{2}+1\right)}{(2 x+5)^{3}}}\right)\]

Applying logarithm properties

We can simplify the right hand side further using logarithm properties. The exponent of a logarithm can be brought out front as a coefficient and the logarithm of a quotient is the difference of the logs of the top and bottom parts. This gives:\[\ln(y) = \frac{1}{5} \left[4\ln(x-3) + \ln(x^2+1) - 3\ln(2x+5)\right]\]

Differentiate both sides

In this step, differentiate both sides of the equation with respect to x. The differentiation of \(\ln(y)\) with respect to x is \(1/y \cdot (dy/dx)\), and for the right-hand side apply regular differentiation rules:\[\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{5} \left[\frac{4}{x-3} + \frac{2x}{x^2+1} - \frac{3\cdot 2}{2x+5}\right]\]

Solve for \(dy / dx\)

In this step, solve the previous equation for \(dy / dx\). Multiplying both sides by y (the original function) gives the derivative of the function:\[\frac{dy}{dx} = y \cdot \frac{1}{5} \left[\frac{4}{x-3} + \frac{2x}{x^2+1} - \frac{3\cdot 2}{2x+5}\right]\]After substituting the original function for y, we get the final derivative of the function:\[\frac{dy}{dx} = \sqrt[5]{\frac{(x-3)^{4}\left(x^{2}+1\right)}{(2 x+5)^{3}}} \cdot \frac{1}{5} \left[\frac{4}{x-3} + \frac{2x}{x^2+1} - \frac{6}{2x+5}\right]\]

Key concepts

Natural Logarithm Properties

Understanding the properties of natural logarithms is crucial for simplifying complex expressions. The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. There are several properties worth noting:

• The logarithm of a product is the sum of the logarithms: \(\ln(ab) = \ln(a) + \ln(b)\).
• The logarithm of a quotient is the difference of the logarithms: \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\).
• The logarithm of a power can be brought out as a coefficient: \(\ln(a^k) = k\cdot\ln(a)\).

These properties enable us to transform the initial complex function into a form that is more manageable for differentiation. For instance, the expression \(\ln\left(\sqrt[5]{\frac{(x-3)^{4}\left(x^{2}+1\right)}{(2 x+5)^{3}}}\right)\) can be simplified using the properties of logarithms to split the radical and the fraction into simpler parts by bringing the power out front and then separating the terms.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with functions that are implicitly defined, meaning y is not isolated. It allows us to differentiate equations directly without solving for y first. In the case of logarithmic differentiation, after taking the natural logarithm of both sides, the equation becomes implicitly defined with respect to y. Instead of isolating y, we differentiate both sides of the equation with respect to x.

Remember that when differentiating a term involving y with respect to x, such as \(\ln(y)\), we apply the chain rule. The derivative of \(\ln(y)\) with respect to y is \(1/y\), but since we are differentiating with respect to x, we multiply by \(dy/dx\) which represents the derivative of y with respect to x. This yields \(\frac{d}{dx}\ln(y) = \frac{1}{y}\cdot\frac{dy}{dx}\). We then use this relationship to find \(\frac{dy}{dx}\) by isolating it on one side of the equation.

Derivative of a Function

The derivative of a function represents the rate at which the function's output changes with respect to a change in its input. It is a fundamental concept in calculus that describes how a quantity varies and is denoted by \(\frac{dy}{dx}\) for function y with respect to x.

In our exercise, after applying logarithmic properties and implicit differentiation, we reach a point where we have an expression for \(\frac{1}{y}\cdot\frac{dy}{dx}\). To isolate the derivative \(\frac{dy}{dx}\), we multiply both sides by the original function \(y\), effectively reversing the effect of taking the logarithm at the beginning. The final step involves substituting the value of \(y\) back into our derivative equation to get the derivative of the function in terms of x alone. This process highlights the importance of understanding the relationship between a function and its derivative, along with the ability to manipulate and simplify expressions to derive the rate of change.