Question

Find $\frac{dy}{dx}$ for $y = \frac{(\cos x).\sqrt{x} + e^{\sec^{-1}x} + (\ln x).\tan x^{2024}}{x^{2023}\tan^{-1}(\ln x) - \sec e^x}$

Answer 1

$\frac{\left(x^{2023}\tan^{-1}(\ln x) - \sec e^x\right) \left(-\sqrt{x}\sin x + \frac{\cos x}{2\sqrt{x}} + \frac{e^{\sec^{-1}x}}{|x|\sqrt{x^2-1}} + \frac{\tan(x^{2024})}{x} + 2024x^{2023}\ln x \sec^2(x^{2024})\right) - \left((\cos x)\sqrt{x} + e^{\sec^{-1}x} + (\ln x)\tan(x^{2024})\right) \left(2023x^{2022}\tan^{-1}(\ln x) + \frac{x^{2022}}{1+(\ln x)^2} - e^x\sec(e^x)\tan(e^x)\right)}{\left(x^{2023}\tan^{-1}(\ln x) - \sec e^x\right)^2}$

Answer 2

To find $\frac{dy}{dx}$ for the given function, we will use the quotient rule for differentiation, which states that if $y = \frac{N(x)}{D(x)}$, then $\frac{dy}{dx} = \frac{D(x) \frac{dN}{dx} - N(x) \frac{dD}{dx}}{(D(x))^2}$.

Let the given function be $y = \frac{N(x)}{D(x)}$, where:

$N(x) = (\cos x)\sqrt{x} + e^{\sec^{-1}x} + (\ln x)\tan(x^{2024})$

$D(x) = x^{2023}\tan^{-1}(\ln x) - \sec(e^x)$

First, we calculate $\frac{dN}{dx}$ by differentiating each term in $N(x)$:

1. Derivative of $(\cos x)\sqrt{x}$: Using the product rule, $\frac{d}{dx}(uv) = u'v + uv'$.

   $\frac{d}{dx}((\cos x)x^{1/2}) = (-\sin x)x^{1/2} + (\cos x)\frac{1}{2}x^{-1/2} = -\sqrt{x}\sin x + \frac{\cos x}{2\sqrt{x}}$

2. Derivative of $e^{\sec^{-1}x}$: Using the chain rule, $\frac{d}{dx}(e^{f(x)}) = e^{f(x)}f'(x)$.

   $\frac{d}{dx}(e^{\sec^{-1}x}) = e^{\sec^{-1}x} \cdot \frac{d}{dx}(\sec^{-1}x) = e^{\sec^{-1}x} \cdot \frac{1}{|x|\sqrt{x^2-1}}$

3. Derivative of $(\ln x)\tan(x^{2024})$: Using the product rule and chain rule.

   $\frac{d}{dx}((\ln x)\tan(x^{2024})) = \frac{d}{dx}(\ln x) \cdot \tan(x^{2024}) + (\ln x) \cdot \frac{d}{dx}(\tan(x^{2024}))$

   $= \frac{1}{x}\tan(x^{2024}) + (\ln x) \cdot \sec^2(x^{2024}) \cdot \frac{d}{dx}(x^{2024})$

   $= \frac{1}{x}\tan(x^{2024}) + (\ln x) \cdot \sec^2(x^{2024}) \cdot (2024x^{2023})$

   $= \frac{\tan(x^{2024})}{x} + 2024x^{2023}\ln x \sec^2(x^{2024})$

Combining these, $\frac{dN}{dx}$: $\frac{dN}{dx} = -\sqrt{x}\sin x + \frac{\cos x}{2\sqrt{x}} + \frac{e^{\sec^{-1}x}}{|x|\sqrt{x^2-1}} + \frac{\tan(x^{2024})}{x} + 2024x^{2023}\ln x \sec^2(x^{2024})$

Next, we calculate $\frac{dD}{dx}$ by differentiating each term in $D(x)$:

1. Derivative of $x^{2023}\tan^{-1}(\ln x)$: Using the product rule and chain rule.

   $\frac{d}{dx}(x^{2023}\tan^{-1}(\ln x)) = \frac{d}{dx}(x^{2023})\tan^{-1}(\ln x) + x^{2023}\frac{d}{dx}(\tan^{-1}(\ln x))$

   $= 2023x^{2022}\tan^{-1}(\ln x) + x^{2023} \cdot \frac{1}{1+(\ln x)^2} \cdot \frac{d}{dx}(\ln x)$

   $= 2023x^{2022}\tan^{-1}(\ln x) + x^{2023} \cdot \frac{1}{1+(\ln x)^2} \cdot \frac{1}{x}$

   $= 2023x^{2022}\tan^{-1}(\ln x) + \frac{x^{2022}}{1+(\ln x)^2}$

2. Derivative of $\sec(e^x)$: Using the chain rule, $\frac{d}{dx}(\sec(f(x))) = \sec(f(x))\tan(f(x))f'(x)$.

   $\frac{d}{dx}(\sec(e^x)) = \sec(e^x)\tan(e^x) \cdot \frac{d}{dx}(e^x) = e^x\sec(e^x)\tan(e^x)$

Combining these, $\frac{dD}{dx}$: $\frac{dD}{dx} = 2023x^{2022}\tan^{-1}(\ln x) + \frac{x^{2022}}{1+(\ln x)^2} - e^x\sec(e^x)\tan(e^x)$

Finally, substitute $N(x)$, $D(x)$, $\frac{dN}{dx}$, and $\frac{dD}{dx}$ into the quotient rule formula: $\frac{dy}{dx} = \frac{\left(x^{2023}\tan^{-1}(\ln x) - \sec e^x\right) \left(-\sqrt{x}\sin x + \frac{\cos x}{2\sqrt{x}} + \frac{e^{\sec^{-1}x}}{|x|\sqrt{x^2-1}} + \frac{\tan(x^{2024})}{x} + 2024x^{2023}\ln x \sec^2(x^{2024})\right) - \left((\cos x)\sqrt{x} + e^{\sec^{-1}x} + (\ln x)\tan(x^{2024})\right) \left(2023x^{2022}\tan^{-1}(\ln x) + \frac{x^{2022}}{1+(\ln x)^2} - e^x\sec(e^x)\tan(e^x)\right)}{\left(x^{2023}\tan^{-1}(\ln x) - \sec e^x\right)^2}$

This is the required derivative $\frac{dy}{dx}$. Due to the extreme complexity of the function, the derivative is also extremely complex and cannot be simplified further.