Problem 49 - Calculus Early Transcendentals

Question

Find the derivatives of the following functions. $$f(v)=\sinh ^{-1} v^{2}$$

Step-by-Step Solution

Answer

Answer: The derivative of the function is \(f'(v) = \frac{2v}{\sqrt{1+v^4}}\).

Step 1: Identify the functions involved and their derivatives

In the given function $$f(v) = \sinh^{-1} v^2 $$We have two functions:$$u(v) = \sinh^{-1} v$$$$v = v^2$$Now, we need to find the derivatives of these functions with respect to \(v\). For the inverse hyperbolic sine function, we know that:$$\frac{d}{dv}(\sinh^{-1}v) = \frac{1}{\sqrt{1+v^2}}$$For the inner function, the derivative is straightforward: $$\frac{d}{dv}(v^2) = 2v$$

Step 2: Apply the chain rule

Now that we have the derivatives of both functions, we can use the chain rule to find the derivative of the composed function. The chain rule states that if \(f(v) = u(v^2)\), then $$\frac{d}{dv}(f(v)) = \frac{d}{dv}(u(v^2))\cdot \frac{d}{dv}(v^2)$$Using the derivatives we found earlier, we have: $$\frac{d}{dv}(f(v)) = \frac{1}{\sqrt{1+(v^2)^2}} \cdot 2v$$

Step 3: Simplify the result

Now, we can simply the expression to arrive at the final solution: $$\frac{d}{dv}(f(v)) = \frac{2v}{\sqrt{1+v^4}}$$So, the derivative of the function $$f(v) = \sinh^{-1} (v^2)$$ is $$f'(v) = \frac{2v}{\sqrt{1+v^4}}$$.