Differentiate tan(cosh(x)) with respect to x.
Understand the Problem
The question is asking to find the derivative of the function tan(cosh(x)) with respect to x. We will use the chain rule and the derivatives of the tan and cosh functions to solve it.
Answer
$$ \frac{dy}{dx} = \sec^2(\cosh(x)) \cdot \sinh(x) $$
Answer for screen readers
The derivative of the function $\tan(\cosh(x))$ with respect to $x$ is: $$ \frac{dy}{dx} = \sec^2(\cosh(x)) \cdot \sinh(x) $$
Steps to Solve
- Identify the function to differentiate
The function we need to differentiate is $y = \tan(\cosh(x))$.
- Apply the chain rule
Since we have an outer function $\tan(u)$ where $u = \cosh(x)$, we use the chain rule: $$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$
- Differentiate the outer function
The derivative of $y = \tan(u)$ with respect to $u$ is: $$ \frac{dy}{du} = \sec^2(u) $$
- Differentiate the inner function
Next, we find the derivative of $u = \cosh(x)$ with respect to $x$: $$ \frac{du}{dx} = \sinh(x) $$
- Combine the derivatives
Now, substitute back: $$ \frac{dy}{dx} = \sec^2(\cosh(x)) \cdot \sinh(x) $$
- Final result
Thus, the derivative of $y = \tan(\cosh(x))$ with respect to $x$ is: $$ \frac{dy}{dx} = \sec^2(\cosh(x)) \cdot \sinh(x) $$
The derivative of the function $\tan(\cosh(x))$ with respect to $x$ is: $$ \frac{dy}{dx} = \sec^2(\cosh(x)) \cdot \sinh(x) $$
More Information
The result combines the derivatives of the tangent and hyperbolic cosine functions using the chain rule. The secant squared function is important in the context of the tangent function, indicating how steep the tangent line is at a given point.
Tips
- Confusing the derivative of tangent with that of sine. Remember, the derivative of $\tan(u)$ is $\sec^2(u)$.
- Forgetting to apply the chain rule properly. Always check if there are functions within functions.
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