Problem 52 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the derivative of the function $$f(u)=\sinh^{-1}(\tan u)$$. Answer: The derivative of the given function $$f(u)=\sinh^{-1}(\tan u)$$ is $$\frac{df(u)}{du} = \frac{\sec^2 u}{\sqrt{1 + (\tan u)^2}}$$.

Step 1: Identify the composite function

We are given the function $$f(u)=\sinh^{-1}(\tan u)$$. Here, the outer function is the inverse hyperbolic sine function $$\sinh^{-1}(x)$$ and the inner function is the tangent function $$\tan u$$.

Step 2: Find the derivative of the outer function

We need to find the derivative of the inverse hyperbolic sine function with respect to $$x$$, $$\frac{d}{dx}\sinh^{-1}(x)$$. The formula for this derivative is: $$\frac{d}{dx}\sinh^{-1}(x) = \frac{1}{\sqrt{1 + x^2}}$$

Step 3: Find the derivative of the inner function

We need to find the derivative of the tangent function with respect to $$u$$, $$\frac{d}{du}\tan u$$. The formula for this derivative is: $$\frac{d}{du}\tan u = \sec^2 u$$

Step 4: Apply the chain rule

Now we will apply the chain rule to find the derivative of the composite function $$f(u)=\sinh^{-1}(\tan u)$$. The chain rule states: $$\frac{df(u)}{du} = \frac{d}{dx}\sinh^{-1}(x) \cdot \frac{d}{du}\left(x\right)$$ where $$x = \tan u$$. The derivatives of the outer and inner functions have been found in Steps 2 and 3. Substituting these derivatives into the chain rule, we get: $$\frac{df(u)}{du} = \frac{1}{\sqrt{1 + x^2}} \cdot \sec^2 u$$

Step 5: Replace $$x$$ with $$\tan u$$

Finally, replace the $$x$$ in the result from the previous step with the inner function $$\tan u$$: $$\frac{df(u)}{du} = \frac{1}{\sqrt{1 + (\tan u)^2}} \cdot \sec^2 u$$

Step 6: Final answer

The derivative of the given function $$f(u)=\sinh^{-1}(\tan u)$$ is: $$\frac{df(u)}{du} = \frac{\sec^2 u}{\sqrt{1 + (\tan u)^2}}$$