Question

Use logarithmic differentiation to find $\frac{dy}{dx}$. $$y=x^{-1/2}{(x^2+3)}^{2/3}(3x-4{)}^4$$

Short answer

The derivative of the function is $\frac{dy}{dx}=x^{-1/2}(x^2+3{)}^{2/3}(3x-4{)}^4(-\frac{1}{2x}+\frac{4x}{3(x^2+3)}+\frac{12}{3x-4})$.

Step-by-step solution

Take the Natural Logarithm of Both Sides

We start by taking the natural logarithm of both sides of the equation, which gives us:$$\ln y=\ln (x^{-1/2}(x^2+3{)}^{2/3}(3x-4{)}^4)$$Utilize the logarithmic property $\ln (abc)=\ln a+\ln b+\ln c$:$$\ln y=\ln (x^{-1/2})+\ln ((x^2+3{)}^{2/3})+\ln ((3x-4{)}^4)$$

Simplify Using Logarithmic Properties

Apply the power rule for logarithms, $\ln (a^b)=b\ln a$, to each term individually:$$\ln y=-\frac{1}{2}\ln x+\frac{2}{3}\ln (x^2+3)+4\ln (3x-4)$$

Differentiate Implicitly with Respect to x

Differentiate both sides of the equation with respect to $x$. The differentiation of $\ln y$ gives $\frac{1}{y}\frac{dy}{dx}$, while each term on the right is differentiated:$$\frac{1}{y}\frac{dy}{dx}=-\frac{1}{2}\frac{1}{x}+\frac{2}{3}\frac{1}{x^2+3}(2x)+4\frac{1}{3x-4}(3)$$This simplifies to:$$\frac{1}{y}\frac{dy}{dx}=-\frac{1}{2x}+\frac{4x}{3(x^2+3)}+\frac{12}{3x-4}$$

Solve for dy/dx

Multiply both sides of the equation by $y$ to isolate $\frac{dy}{dx}$:$$\frac{dy}{dx}=y(-\frac{1}{2x}+\frac{4x}{3(x^2+3)}+\frac{12}{3x-4})$$Substitute back the expression for $y$:$$\frac{dy}{dx}=x^{-1/2}(x^2+3{)}^{2/3}(3x-4{)}^4(-\frac{1}{2x}+\frac{4x}{3(x^2+3)}+\frac{12}{3x-4})$$

Final Expression for dy/dx

This is the final expression for the derivative of $y$ with respect to $x$ using logarithmic differentiation.$$\frac{dy}{dx}=x^{-1/2}(x^2+3{)}^{2/3}(3x-4{)}^4(-\frac{1}{2x}+\frac{4x}{3(x^2+3)}+\frac{12}{3x-4})$$

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with functions that are not explicitly given as one variable in terms of another, or when equations involve products or compositions of functions. In such cases, finding the derivative directly can be complex. Instead, we differentiate both sides of an equation with respect to a variable, usually $x$, while treating other variables, such as $y$, as implicit functions of $x$.

For example, if we have an equation $y=f(x)$, instead of directly solving for $y$ and then differentiating, we approach it by taking the derivative of each side concerning $x$. This often involves applying the chain rule, which helps handle the derivative of $y$ as $\frac{dy}{dx}$.

In our exercise, when applying implicit differentiation to $\ln y$, we treat $y$ as an implicit function of $x$. Thus, the derivative of $\ln y$ with respect to $x$ is $\frac{1}{y}\cdot \frac{dy}{dx}$, leveraging the chain rule. This allows us to find $\frac{dy}{dx}$ by addressing each term of the simplified logarithmic expression derived from $\ln (abc)=\ln a+\ln b+\ln c$.

Natural Logarithm

The natural logarithm, denoted as $\ln (x)$, is a specific logarithm with the base $e$, where $e\approx 2.718$. It appears frequently in calculus due to its properties that simplify differentiation and integration processes. The natural logarithm is particularly useful for functions involving exponential growth or decay.

When using natural logarithms in differentiation, it transforms multiplicative complexities into additive simple terms. For example, if you have a product of functions $f(x)\times g(x)$, the logarithmic property $\ln (fg)=\ln (f)+\ln (g)$ lets us differentiate individually, making the overall process simpler.

In the context of logarithmic differentiation, natural logarithms help manage expression complexities by applying properties like $\ln (a^b)=b\ln a$. This property is handy when dealing with powers, as in our exercise, where expressions such as $(x^2+3{)}^{2/3}$ are reduced to $\frac{2}{3}\ln (x^2+3)$. This reduces differentiation to handling easier terms and leads to straight-forward derivative calculations.

Power Rule for Logarithms

The power rule for logarithms is an essential logarithmic property that results from the general rule $\ln (a^b)=b\ln a$. It allows us to "move" the exponent in front of the logarithm, simplifying the differentiation or integration of expressions involving powers.

This property is particularly useful in logarithmic differentiation because it can break down seemingly complex problems into manageable pieces. By converting powers into coefficients, we simplify the differentiation process.

Consider our main exercise: initially, we have expressions such as $x^{-1/2}$, which can be rewritten using the power rule as $-\frac{1}{2}\ln x$. This application across our terms $(x^2+3{)}^{2/3}$ and $(3x-4{)}^4$ as seen in the step-by-step solution lets us create a summable form where each term becoming a straightforward function of $x$.

With simplified terms, implicit differentiation can be applied effectively, making it possible to find derivatives of complex algebraic expressions with ease.