Problem 43
Question
Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. \(y=(\sin x)^{x}\)
Step-by-Step Solution
Verified Answer
The derivative is \( \frac{dy}{dx} = (\sin x)^x (\ln(\sin x) + x \cot x) \).
1Step 1: Take the Natural Logarithm
To use logarithmic differentiation, we first take the natural logarithm on both sides of the equation. This will help to simplify the power expression.\[ \ln y = \ln \, ((\sin x)^x) \]
2Step 2: Use Logarithm Properties
Apply the power rule of logarithms, which states that \(\ln(a^b) = b\ln(a)\). So, we can rewrite the right side of the equation:\[ \ln y = x \ln (\sin x) \]
3Step 3: Differentiate Both Sides with Respect to x
Differentiate both sides of the equation with respect to \(x\). Note that on the left side, we use implicit differentiation:\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (x \ln (\sin x)) \]
4Step 4: Differentiate the Right Side using Product Rule
Use the product rule on the right side, given by \((u \cdot v)' = u'v + uv'\), where \(u = x\) and \(v = \ln (\sin x)\).\[ \frac{d}{dx}(x \ln (\sin x)) = 1 \cdot \ln (\sin x) + x \cdot \left(\frac{1}{\sin x}\right) \cdot \cos x \] Which simplifies to:\[ \ln (\sin x) + x \cot x \]
5Step 5: Solve for \(\frac{dy}{dx}\)
Multiply both sides of the equation by \(y\) to isolate \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = y \left( \ln (\sin x) + x \cot x \right) \]
6Step 6: Substitute y Back into the Equation
Since we know that \(y = (\sin x)^x\), substitute it back into the equation:\[ \frac{dy}{dx} = (\sin x)^x \left( \ln (\sin x) + x \cot x \right) \]
Key Concepts
Power ruleProduct ruleImplicit differentiationNatural logarithm
Power rule
The power rule is a basic principle in calculus used to differentiate functions of the form \(x^n\), where \(n\) is a constant. It states that if \(y = x^n\), then the derivative \(\frac{dy}{dx} = nx^{n-1}\). However, in logarithmic differentiation, the power rule is applied in a slightly different way.
When we have a function like \((\sin x)^x\), we encounter a base that is a function of \(x\) raised to the power of another function of \(x\). By taking the natural logarithm, we use the logarithmic property that allows the power to come out as a coefficient, transforming the problem into one that is easier to handle.
This manipulation is essential in cases where both the base and the exponent depend on the same variable, making traditional power rule application infeasible without logarithmic differentiation.
When we have a function like \((\sin x)^x\), we encounter a base that is a function of \(x\) raised to the power of another function of \(x\). By taking the natural logarithm, we use the logarithmic property that allows the power to come out as a coefficient, transforming the problem into one that is easier to handle.
This manipulation is essential in cases where both the base and the exponent depend on the same variable, making traditional power rule application infeasible without logarithmic differentiation.
Product rule
The product rule is a vital tool in differentiation used when dealing with the derivative of two multiplicative functions, \(u(x)\) and \(v(x)\). The rule is defined as:
In our problem, after taking the natural logarithm and applying the power rule for logarithms, we identify two separate functions to apply the product rule. These functions are \(u = x\) and \(v = \ln(\sin x)\).
Differentiating \(u\) gives 1, since the derivative of \(x\) with respect to \(x\) is 1. To differentiate \(v\), we use the chain rule because \(\ln(\sin x)\) involves a composition of functions. The result \(\frac{1}{\sin x} \cdot \cos x\) facilitates the application of the product rule, yielding \(\ln(\sin x) + x\cot x\) as part of the solution.
- If \(y = u(x)v(x)\), then the derivative \(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\).
In our problem, after taking the natural logarithm and applying the power rule for logarithms, we identify two separate functions to apply the product rule. These functions are \(u = x\) and \(v = \ln(\sin x)\).
Differentiating \(u\) gives 1, since the derivative of \(x\) with respect to \(x\) is 1. To differentiate \(v\), we use the chain rule because \(\ln(\sin x)\) involves a composition of functions. The result \(\frac{1}{\sin x} \cdot \cos x\) facilitates the application of the product rule, yielding \(\ln(\sin x) + x\cot x\) as part of the solution.
Implicit differentiation
Implicit differentiation is used when a function is not explicitly solved for one variable. Instead, both sides of the equation may involve both variables in a complex relation.
In this exercise, after using the natural logarithm, we obtain an equation involving \(y\) and \(x\), specifically \(\ln y = x \ln(\sin x)\). Here, \(y\) is defined implicitly as a function of \(x\). To find \(\frac{dy}{dx}\), implicit differentiation is necessary.
When differentiating \(\ln y\) with respect to \(x\), we treat \(y\) as a function of \(x\), leading to \(\frac{1}{y} \cdot \frac{dy}{dx}\). This technique allows us to differentiate equations where the dependent variable cannot be easily isolated.
Without implicit differentiation, tackling the derivative of complex combinations like the one in this problem would be much more challenging, if not impossible in some instances.
In this exercise, after using the natural logarithm, we obtain an equation involving \(y\) and \(x\), specifically \(\ln y = x \ln(\sin x)\). Here, \(y\) is defined implicitly as a function of \(x\). To find \(\frac{dy}{dx}\), implicit differentiation is necessary.
When differentiating \(\ln y\) with respect to \(x\), we treat \(y\) as a function of \(x\), leading to \(\frac{1}{y} \cdot \frac{dy}{dx}\). This technique allows us to differentiate equations where the dependent variable cannot be easily isolated.
Without implicit differentiation, tackling the derivative of complex combinations like the one in this problem would be much more challenging, if not impossible in some instances.
Natural logarithm
The natural logarithm, denoted as \(\ln\), is a logarithm with base \(e\), where \(e \approx 2.71828\). It's frequently used in calculus due to its simplifying properties, particularly in differentiation.
In the context of logarithmic differentiation, the natural logarithm is essential because it allows us to simplify expressions by applying logarithmic rules.
One major property used is \(\ln(a^b) = b\ln(a)\), which helps transform intricate power functions into a more manageable linear form. By converting the original function \((\sin x)^x\) into \(x \ln(\sin x)\), we can more readily apply standard differentiation techniques like the product rule.
The natural logarithm not only reduces complexity but also enables us to handle derivatives of functions where both the base and the exponent simultaneously depend on the variable of differentiation.
In the context of logarithmic differentiation, the natural logarithm is essential because it allows us to simplify expressions by applying logarithmic rules.
One major property used is \(\ln(a^b) = b\ln(a)\), which helps transform intricate power functions into a more manageable linear form. By converting the original function \((\sin x)^x\) into \(x \ln(\sin x)\), we can more readily apply standard differentiation techniques like the product rule.
The natural logarithm not only reduces complexity but also enables us to handle derivatives of functions where both the base and the exponent simultaneously depend on the variable of differentiation.
Other exercises in this chapter
Problem 43
Evaluate the integrals in Exercises \(41-62\). $$ \int_{\ln 2}^{\ln 3} e^{x} d x $$
View solution Problem 43
Evaluate the integrals in Exercises \(37-54\). $$ \int_{1}^{2} \frac{2 \ln x}{x} d x $$
View solution Problem 44
Evaluate the integrals in Exercises \(41-50\) . $$ \int 4 \cosh (3 x-\ln 2) d x $$
View solution Problem 44
Find the limits in Exercises \(41-48 .\) (If in doubt, look at the function's graph.) $$ \lim _{x \rightarrow-\infty} \tan ^{-1} x $$
View solution