Problem 47 - Calculus Early Transcendentals

Question

Find the derivatives of the following functions. $$f(x)=\cosh ^{-1} 4 x$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(f(x) = \cosh^{-1}(4x)\). Solution: Using the chain rule and the derivative formula for the inverse hyperbolic cosine function, we find that the derivative of the given function is: $$\frac{d}{dx}f(x) = \frac{4}{\sqrt{16x^2 - 1}}$$

Step 1: Recall the derivative formula for the inverse hyperbolic cosine

Recall that the derivative of the inverse hyperbolic cosine function with respect to \(x\) is given by: $$\frac{d}{dx}\cosh^{-1}(x) = \frac{1}{\sqrt{x^2 - 1}}$$ We will need this formula to find the derivative of \(f(x)\).

Step 2: Apply the chain rule

Since our function \(f(x)\) is a composition of two functions (\(\cosh^{-1}(u)\) with \(u=4x\)), we need to use the chain rule to find its derivative. The chain rule states that: $$\frac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$$ In our case, \(f(u) = \cosh^{-1}(u)\) and \(g(x) = 4x\).

Step 3: Compute the derivatives of each function

First, we need to compute the derivative of \(f(u) = \cosh^{-1}(u)\) with respect to \(u\) using the formula mentioned in Step 1: $$\frac{d}{du}\cosh^{-1}(u) = \frac{1}{\sqrt{u^2 - 1}}$$ Next, we need to compute the derivative of \(g(x) = 4x\) with respect to \(x\): $$\frac{d}{dx}(4x) = 4$$

Step 4: Apply the chain rule

Now we can apply the chain rule to find the derivative of \(f(x) = \cosh^{-1}(4x)\): $$\frac{d}{dx}\cosh^{-1}(4x) = \frac{d}{du}\cosh^{-1}(4x) \cdot \frac{d}{dx}(4x)$$ Substituting the derivatives of each function from Step 3, we get: $$\frac{d}{dx}\cosh^{-1}(4x)= \frac{1}{\sqrt{(4x)^2 - 1}}\cdot 4$$

Step 5: Simplify the derivative

Simplify the expression to get the final answer for the derivative of \(f(x)\): $$\frac{d}{dx}\cosh^{-1}(4x) = \frac{4}{\sqrt{(4x)^2 - 1}} = \frac{4}{\sqrt{16x^2 - 1}}$$ The derivative of the given function is: $$\frac{d}{dx}f(x) = \frac{4}{\sqrt{16x^2 - 1}}$$