Question

Find the derivatives of the following functions. $$f(x)={\mathrm{csch}}^{-1}\left(\frac{2}{x}\right)$$

Short answer

Question: Find the derivative of the function $$f(x)={\mathrm{csch}}^{-1}\left(\frac{2}{x}\right)$$ Answer: The derivative of the given function is $$f^{\prime }(x)=\frac{1}{x\sqrt{\frac{x^2+4}{x^2}}}$$

Step-by-step solution

Recall the derivative of inverse hyperbolic cosecant

Recall that the derivative of the inverse hyperbolic cosecant ${\mathrm{csch}}^{-1}(u)$ with respect to $u$ is $$\frac{d}{du}{\mathrm{csch}}^{-1}(u)=-\frac{1}{u\sqrt{1+u^2}}$$ We will use this result and apply the chain rule to find the derivative of the given function.

Apply the chain rule

Let $g(x)=\frac{2}{x}$. Then our function can be written as $$f(x)={\mathrm{csch}}^{-1}(g(x))$$ Applying the chain rule to find the derivative of $f(x)$ with respect to $x$, we get $$f^{\prime }(x)=\frac{d({\mathrm{csch}}^{-1}(g(x)))}{dx}=\frac{d({\mathrm{csch}}^{-1}(g(x)))}{dg(x)}\cdot \frac{dg(x)}{dx}$$

Differentiate $g(x)$ and apply the derivative of inverse hyperbolic cosecant

First, differentiate $g(x)$ with respect to $x$: $$\frac{dg(x)}{dx}=\frac{d(2/x)}{dx}=-\frac{2}{x^2}$$ Using the result from Step 1, differentiate ${\mathrm{csch}}^{-1}(g(x))$ with respect to $g(x)$: $$\frac{d({\mathrm{csch}}^{-1}(g(x)))}{dg(x)}=-\frac{1}{g(x)\sqrt{1+g(x{)}^2}}=-\frac{1}{\frac{2}{x}\sqrt{1+{\left(\frac{2}{x}\right)}^2}}$$

Combine the derivatives and simplify the expression

Now combine the derivatives according to the chain rule: $$f^{\prime }(x)=-\frac{1}{\frac{2}{x}\sqrt{1+{\left(\frac{2}{x}\right)}^2}}\cdot -\frac{2}{x^2}$$ To simplify this expression, we can multiply the numerators and denominators: $$f^{\prime }(x)=\frac{2}{x^2}\cdot \frac{x}{2\sqrt{1+\frac{4}{x^2}}}$$ Now, cancel out the common factors of $x$ and $2$: $$f^{\prime }(x)=\frac{1}{x\sqrt{1+\frac{4}{x^2}}}$$ Finally, we can multiply the denominator by $\frac{x^2}{x^2}$ to remove the fraction inside the square root: $$f^{\prime }(x)=\frac{1}{x\sqrt{\frac{x^2+4}{x^2}}}$$ So, the derivative of the given function is: $$f^{\prime }(x)=\frac{1}{x\sqrt{\frac{x^2+4}{x^2}}}$$

Key concepts

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus for differentiating composed functions, where one function is nested within another. To understand this concept, imagine two functions, one inside the other, like a set of Russian dolls. In its core, the chain rule tells us that to find the derivative of such composite functions, you take the derivative of the outer function with respect to the inner function and then multiply it by the derivative of the inner function with respect to the variable.

For instance, if you have a function like $h(x)=f(g(x))$, the derivative $h^{\prime }(x)$ is found by $h^{\prime }(x)=f^{\prime }(g(x))\times g^{\prime }(x)$. It's a straightforward yet powerful tool that allows for the calculation of derivatives that may not be immediately obvious. The chain rule's versatility makes it an indispensable technique in calculus, especially when dealing with complex functions.

Differentiating Inverse Hyperbolic Functions

Handling inverse hyperbolic functions such as ${\mathrm{csch}}^{-1}(x)$ or ${\sinh }^{-1}(x)$ means understanding their derivatives. These functions are the inverses of hyperbolic functions and they require specific formulas for finding their derivatives. For example, the inverse hyperbolic cosecant, ${\mathrm{csch}}^{-1}(u)$, has a derivative given by $-\frac{1}{u\sqrt{1+u^2}}$.

Practical Use

These formulas are not just for academic exercises; they are used in fields that model rapid growth or decay, heat transfer, and relativity, among others. When a student learns to differentiate these functions, they're not just solving a math problem; they're learning a concept that has real-world applications.

Calculus Problem Solving

Approaching calculus problems involves a blend of understanding theoretical concepts and applying them methodically. Considering the example of finding the derivative of ${\mathrm{csch}}^{-1}\left(\frac{2}{x}\right)$, we break down the problem into manageable steps.

Start by recognizing the underlying structure of the function, which in this case involves an inverse hyperbolic function. Next, it's crucial to recall the specific derivative formula that applies to that function. Then, break down the problem using the chain rule and carefully work through each step, ensuring that each derivative component is accurately determined. Simplifying the expression at the end can often involve algebraic manipulation, such as rationalizing the denominator or combining like terms. Details matter, and each step lays the foundation for the next—ensuring thorough understanding at each stage is key to mastering calculus problem solving.

Applying the Chain Rule

In practice, applying the chain rule to a function like ${\mathrm{csch}}^{-1}\left(\frac{2}{x}\right)$ extends beyond merely memorizing a procedure; it requires an understanding of the function's components. Breaking down the function into its parts: $f(x)={\mathrm{csch}}^{-1}(g(x))$ and $g(x)=\frac{2}{x}$, demonstrates that $g(x)$ is nested inside the inverse hyperbolic function.

Once we have identified this composition, we apply the derivative of the outer function (${\mathrm{csch}}^{-1}(u)$) to the inner function ($g(x)$), then multiply by the derivative of the inner function ($g^{\prime }(x)$). This step-by-step multiplicative process is at the heart of the chain rule's application and allows for the correct and efficient calculation of complex derivatives.