Question

Critical points Find the critical points of the function $f(x)={\sinh }^{2}x\cosh x$

Short answer

Question: Determine the critical points of the function $f(x)={\sinh }^{2}x\cosh x$. Answer: The critical points are given by $x=n\pi$ for any integer $n$.

Step-by-step solution

Find the first derivative f'(x)

To find the first derivative, we will apply the product rule, which states that for two differentiable functions u(x) and v(x), their product's derivative is given by $d(uv)/dx=udv/dx+vdu/dx$. Let $u(x)={\sinh }^{2}x$ and $v(x)=\cosh x$. Then we have: $f^{\prime }(x)=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$ Now we need to find the derivatives of u(x) and v(x). For $u(x)={\sinh }^{2}x$, apply the chain rule: $({\sinh }^{2}x{)}^{\prime }=2\sinh x(\cosh x)$. For $v(x)=\cosh x$, the derivative is: $(\cosh x{)}^{\prime }=\sinh x$. Now plug the derivatives back into the expression for the first derivative: $f^{\prime }(x)=2\sinh x(\cosh x)\cdot \cosh x+{\sinh }^{2}x\cdot \sinh x$

Solve for critical points

To find the critical points of f(x), we need to solve the equation $f^{\prime }(x)=0$. We have: $2\sinh x(\cosh x)\cdot \cosh x+{\sinh }^{2}x\cdot \sinh x=0$ Factor out a $\sinh x$: $\sinh x(2{\cosh }^{2}x\cosh x+{\sinh }^{2}x)=0$ The critical points occur when this equation is equal to 0. There will be a solution when $\sinh x=0$. The $\sinh x$ function is equal to 0 whenever $x$ is an integer multiple of $\pi$. Therefore, the critical points are: $x=n\pi$, where $n$ is an integer. The other term, $2{\cosh }^{2}x\cosh x+{\sinh }^{2}x$, does not have any solutions that would result in the derivative being 0 or undefined. Therefore, we have found all the critical points of the function $f(x)={\sinh }^{2}x\cosh x$. The critical points are given by $x=n\pi$ for any integer $n$.

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions, but instead of being based on the unit circle, they are based on a hyperbola. The main hyperbolic functions include the hyperbolic sine ($\sinh (x)$) and hyperbolic cosine ($\cosh (x)$). These functions are essential in various mathematical fields, including calculus and complex analysis.

• The hyperbolic sine is defined as: $\sinh (x)={\displaystyle \frac{e^x-e^{-x}}{2}}$
• The hyperbolic cosine is defined as: $\cosh (x)={\displaystyle \frac{e^x+e^{-x}}{2}}$

Hyperbolic functions often arise in calculus when dealing with exponential growth processes. They have similar properties to trigonometric functions, such as periodicity and symmetry, but differ in their algebraic forms and derivatives.

Derivative

Derivatives are a fundamental concept in calculus that allow us to understand how a function changes as its input changes. The derivative of a function at a given point provides the slope of the tangent line to the graph of the function at that point.For the function given, $f(x)={\sinh }^{2}x\cosh x$, the goal is to find $f^{\prime }(x)$, which means finding the rate at which$f(x)$ changes with respect to $x$. The process involves applying rules such as the product rule and chain rule to differentiate the function step by step.Steps to find the derivative include:

• Identifying function components that need differentiating.
• Applying the appropriate differentiation rules.
• Combining the results to find $f^{\prime }(x)$.

Understanding derivatives and these rules allow us to solve problems involving rates of change, which is crucial for many real-world applications.

Product Rule

The product rule is a technique in calculus used to find the derivative of the product of two functions. When you have two functions, say $u(x)$ and $v(x)$, the product rule states that the derivative of their product $u(x)v(x)$ is given by: $${\displaystyle \frac{d}{dx}}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$In the context of the function $f(x)={\sinh }^{2}x\cosh x$, let:

• $u(x)={\sinh }^{2}x$
• $v(x)=\cosh x$

By applying the product rule, it's clear that you must differentiate each function separately, calculate their derivatives, and then plug these results into the product rule formula. This results in the derivative $f^{\prime }(x)=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$, which simplifies the process of differentiating the entire function.

Chain Rule

The chain rule is a method used in calculus to differentiate composite functions. This rule simplifies the process of finding derivatives when functions are nested inside each other. The chain rule states:If a variable $y$ depends on $u$, which itself depends on $x$, then: $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{du}}\cdot {\displaystyle \frac{du}{dx}}$$For a precise example in the exercise with $u(x)={\sinh }^{2}x$:

• Recognize that ${\sinh }^{2}x$ is a composite function because it involves $(\sinh (x){)}^2$.
• First find the derivative of the inner function $\sinh (x)$, which is $\cosh (x)$.
• Then multiply by the derivative of the outer function, which is generally $2x$ for a squared term.

Thus, by applying the chain rule, we obtain that $({\sinh }^{2}x{)}^{\prime }=2\sinh x\cosh x$. This is essential for simplifying the differentiation process of more complex functions and is used when deriving $f^{\prime }(x)$ in the given problem.